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STRENGTH  OF   MATERIALS 

A  TEXT-BOOK 

FOR 

SECONDARY  TECHNICAL  SCHOOLS 


BY 

MANSFIELD  MERRIMAN 

MEMBER  or  INTERNATIONAL  ASSOCIATION  FOR  TESTING  MATERIALS 


SIXTH  EDITION,  REVISED  AND  RESET 
TOTAL  ISSUE,  TWENTY-TWO  THOUSAND 


NEW  YORK 

JOHN  WILEY  &  SONS,  INC. 

LONDON:  CHAPMAN  &  HALL,  LIMBED 

1915 
*f  O  +  O  O 

MAY  3       1917 


Copyright,  1897,  1906,  1912 

BY 

MANSFIELD  MERRIMAN 


First  Edition,  November,  1897;  reprinted,  1898 

Second  Edition,  December,  1898;  reprinted,  1899 

Third  Edition,  November,  1900;  reprinted,  1901,  1903 

Fourth  Edition,  Revised,  October,  1905 
Fifth  Edition,  Enlarged,  September,  1906;  reprinted,  1907,  1908. 

1909,  1910 

Sixth  Edition,  Revised  and  Reset,  July,  1912 
Reprinted,  December,  1912;  August,  1913;  April,  1915 


;  OF  THE  PUBLISHERS  PRINTING  COMPANY,  NEW  YORK.  U.  S.  A. 


I      A 

^  0  5 
55  3 


PEEFAOE   TO  THE  FIRST  EDITION 

IN  the  following  pages  the  attempt  is  made  to  give  a 
presentation  of  the  subject  of  the  strength  of  materials, 
beams,  columns,  and  shafts,  which  may  be  understood 
by  those  not  acquainted  with  the  calculus.  The  degree 
of  mathematical  preparation  required  is  merely  that 
now  given  in  high  schools,  and  includes  only  arithmetic, 
algebra,  geometry,  and  such  a  course  in  mechanics  as 
is  found  in  elementary  works  on  physics.  In  particular 
the  author  has  had  in  mind  the  students  in  the  higher 
classes  of  manual  training  schools,  and  it  has  been  his 
aim  to  present  the  subject  in  such  an  elementary  manner 
that  it  may  be  readily  comprehended  by  them  and  at 
the  same  time  cover  all  the  essential  principles  and 
methods. 

As  the  title  implies  the  book  deals  mainly  with  ques- 
tions of  strength,  the  subject  of  elastic  deformations 
occupying  a  subordinate  place.  As  the  deductions  of  the 
deflections  of  beams  are  best  made  by  the  calculus  they 
are  not  here  attempted,  but  the  results  are  stated  so 
that  the  student  may  learn  their  uses;  later,  if  he  con- 
tinues the  study  of  engineering,  his  appreciation  of 
the  proofs  that  he  will  then  read  will  be  accompanied 
with  true  scientific  interest. 

All  the  rules  for  the  investigation  and  design  of  com- 
mon beams,  including  the  subject  of  moment  of  inertia, 
are  here  presented  by  simple  algebraic  and  geometric 


iV  PREFACE 

methods.  As  the  mechanical  ideas  involved  are  by  far 
the  most  difficult  part  of  the  subject,  a  special  effort  has 
been  made  to  clearly  present  them,  and  to  illustrate 
them  by  numerous  practical  numerical  examples. 

A  chapter  on  the  manufacture  and  general  properties 
of  materials  is  given,  as  also  one  on  resilience  and  impact. 
Problems  for  students  to  solve  are  presented,  and  it 
should  be  strongly  insisted  upon  that  these  should  be 
thoroughly  and  completely  worked  out.  It  is  indeed 
only  by  the  solution  of  many  numerical  exercises  that 
a  good  knowledge  of  the  theory  of  the  subject  can  be 
acquired. 

1TOTE  TO  THE  SIXTH  EDITION 

In  the  fifth  edition  a  new  chapter  was  added  on  rein- 
forced concrete,  especially  columns  and  beams.  In  this 
edition  a  new  chapter  on  combined  stresses  is  added, 
numerous  changes  have  been  made  throughout,  and  many 
new  problems  introduced.  The  number  of  articles  is 
increased  from  72  to  91,  the  number  of  cuts  from  48  to 
54,  and  the  number  of  problems  from  140  to  230.  For 
45  of  the  new  problems  my  thanks  are  due  to  Professor 
J.  M.  Jameson,  of  the  Pratt  Institute.  It  is  hoped  that 
the  volume  in  its  new  form  may  advance  the  cause  of  sound 
technical  education  more  effectively  than  before. 

MANSFIELD  MERRIMAN. 

NEW  YORK,  JUNE,  1912. 


CONTENTS 

CHAP.  1.     ELASTIC  AND  ULTIMATE  STRENGTH 

PAGE 

ART.  1.  DIRECT  STRESSES.  2.  THE  ELASTIC  LIMIT.  3.  ULTI- 
MATE STRENGTH.  4.  TENSION.  5.  COMPRESSION.  6.  SHEAR. 
7.  WORKING  UNIT-STRESSES.  8.  REVIEW  PROBLEMS 1 

CHAP.  2.     GENERAL  PROPERTIES 

ART.  9.  AVERAGE  WEIGHTS.  10.  TESTING  MACHINES.  11. 
TIMBER.  12.  BRICK.  13.  STONE.  14.  CAST  IRON.  15.  WROUGHT 
IRON.  16.  STEEL.  17.  OTHER  MATERIALS.  18.  REVIEW  PROB- 
LEMS    17 

CHAP.  3.     MOMENTS   FOR   BEAMS 

ART.  19.  THE  PRINCIPLE  OF  MOMENTS.  20.  REACTION'S  FOR 
SUPPORTS.  21.  BENDING  MOMENT.  22.  RESISTING  MOMENTS. 
23.  CENTERS  OF  GRAVITY.  24.  MOMENTS  OF  INERTIA.  25. 
REVIEW  PROBLEMS 37 

CHAP.  4.     CANTILEVER  AND  SIMPLE   BEAMS 

ART.  26.  DEFINITIONS  AND  PRINCIPLES.  27.  RESISTANCE  TO 
SHEARING.  28.  RESISTANCE  TO  BENDING.  29.  SAFE  LOADS  FOR 
BEAMS.  30.  INVESTIGATION  OF  BEAMS.  31.  DESIGN  OF  BEAMS. 
32.  COMPARATIVE  STRENGTHS.  33.  STEEL  I  BEAMS.  34. 
BEAMS  OF  UNIFORM  STRENGTH.  35.  REVIEW  PROBLEMS  ....  56 

CHAP.  5.     COLUMNS   OR  STRUTS 

ART.  36.  GENERAL  PRINCIPLES.  37.  RADIUS  OF  GYRATION. 
38.  FORMULA  FOR  COLUMNS.  39.  SAFE  LOADS  FOR  COLUMNS. 
40.  INVESTIGATION  OF  COLUMNS.  41.  DESIGN  OF  COLUMNS. 
42.  ECCENTRIC  LOADS.  43.  THE  STRAIGHT-LINE  FORMULA.  44. 
REVIEW  PROBLEMS .  77 

CHAP.  6.     THE   TORSION   OF  SHAFTS 

ART.  45.  PHENOMENA  OF  TORSION.  46.  POLAR  MOMENT  OF 
INERTIA.  47.  FORMULA  FOR  TORSION.  48.  SHAFTS  TO  TRANS- 
MIT POWER.  49.  SOLID  SHAFTS.  50.  HOLLOW  SHAFTS.  51. 
REVIEW  PROBLEMS 92 


Vl  CONTENTS 

CHAP.  7.     ELASTIC   DEFORMATIONS 

PAGE 

ART.  52.  THE  COEFFICIENT  OF  ELASTICITY.  53.  ELONGATION 
UNDER  TENSION.  54.  SHORTENING  UNDER  COMPRESSION.  55. 
DEFLECTION  OF  CANTILEVER  BEAM.S.  56.  DEFLECTION  OF  SIM- 
PLE BEAMS.  57.  RESTRAINED  BEAMS.  58.  TWIST  IN  SHAFTS. 
59.  REVIEW  PROBLEMS 103 

CHAP.  8.     MISCELLANEOUS   APPLICATIONS 

ART.  60.  WATER  AND  STEAM  PIPES.  61.  RIVETED  LAP 
JOINTS.  62.  RIVETED  BUTT  JOINTS.  63.  STRESSES  DUE  TO 
TEMPERATURE.  64.  SHRINKAGE  OF  HOOPS.  65.  SHAFT  COUP- 
LINGS. 66.  RUPTURE  OF  BEAMS  AND  SHAFTS.  67.  REVIEW 
PROBLEMS 114 

CHAP.  9.     REINFORCED   CONCRETE 

ART.  68.  CONCRETE  AND  STEEL.  69.  COMPOUND  BARS.  70. 
SHORT  COLUMNS.  71.  BEAMS  WITH  SYMMETRIC  REINFORCE- 
MENT. 72.  UNSYMMETRIC  REINFORCEMENT.  73.  DESIGN  OF 
BEAMS.  74.  GENERAL  DISCUSSIONS.  75.  REVIEW  PROBLEMS  . .  126 

CHAP.  10.     COMBINED   STRESSES 

ART.  76.  COMPRESSION  AND  FLEXURE.  77.  TENSION  AND 
FLEXURE.  78.  SHEAR  AND  AXIAL  STRESS.  79.  COMPRESSION 
AND  TORSION.  80.  FLEXURE  AND  TORSION.  81.  TENSION  AND 
COMPRESSION.  82.  REVIEW  PROBLEMS 146 

CHAP.  11.     RESILIENCE   OF   MATERIALS 

ART.  83.  FUNDAMENTAL  IDEAS.  84.  ELASTIC  RESILIENCE  OF 
BARS.  85.  ELASTIC  RESILIENCE  OF  BEAMS.  86.  ULTIMATE 
RESILIENCE.  87.  SUDDEN  LOADS.  88.  STRESSES  DUE  TO  IM- 
PACT. 89.  REPEATED  STRESSES.  90.  REVIEW  PROBLEMS.  91. 
ANSWERS  TO  PROBLEMS 154 

INDEX..  .   167 


STRENGTH  OF  MATERIALS 

CHAPTER  1 

ELASTIC  AND  ULTIMATE    STRENGTH 
ART.  1.    DIRECT  STRESSES 

A  'stress'  is  an  internal  resistance  which  balances  an 
exterior  force.  When  a  weight  of  500  pounds  is  suspended 
by  a  rope,  a  stress  of  500  pounds  exists  in  every  cross- 
section  of  the  rope;  or,  if  this  rope  is  cut  anywhere  and 
the  ends  are  connected  by  a  spring  balance,  this  will 
register  500  pounds.  Stresses  are  measured  in  the  same 
units  as  forces,  namely,  in  pounds,  tons,  or  kilograms. 

A  'unit-stress'  is  the  stress  on  a  unit  of  area;  this  is 
expressed  in  pounds  per  square  inch  or  in  kilograms  per 
square  centimeter.  Thus,  when  a  bar  of  three  square 
inches  in  cross-section  is  subject  to  a  pull  of  12  000  pounds, 
the  unit-stress  is  4000  pounds  per  square  inch  for  the 
usual  case  when  the  total  stress  is  uniformly  distributed 
over  the  cross-section. 

Three  kinds  of  direct  stress  are  produced  by  exterior 
forces  which  act  on  a  body  in  such  a  way  as  to  tend  to 
change  its  shape;  these  are, 

Tension,  tending  to  pull  apart,  as  in  a  rope. 

Compression,  tending  to  push  together,  as  in  a  wall  or  column. 

Shear,  tending  to  cut  across,  as  in  punching  a  plate. 

The  forces  which  produce  these  kinds  of  stress  may  be 
called  tensile,  compressive,  and  shearing  forces,  while  the 
i  1 


2  ELASTIC  AND    ULTIMATE   STRENGTH  CH.  1 

stresses  themselves  are  frequently  called  tensile,   com- 
pressive,  and  shearing  stresses. 

A  stress  is  always  accompanied  by  a  'deformation'  or 
change  of  shape  of  the  body.  As  the  applied  force  increases 
the  deformation  and  the  stress  likewise  increase,  and  if 
the  force  is  large  enough  it  finally  overcomes  the  stress 
and  the  rupture  of  the  body  follows. 

Tension  and  compression  differ  only  in  regard  to  direc- 
tion (Fig.  1).  A  tensile  stress  in  a  bar  occurs  when  two 
forces  of  equal  intensity  act  upon  its  ends,  each  acting 


Fig.l 

away  from  the  end  of  the  bar.  In  compression  the  direc- 
tion of  the  forces  is  reversed,  each  acting  toward  the  end 
of  the  bar.  The  tensile  force  produces  a  deformation 
called  'elongation'  and  the  compressive  force  produces 
a  deformation  called  'shortening.'  If  P  is  the  force  in 
pounds,  then  the  total  stress  in  every  section  of  the  bar 
is  equal  to  P. 

Shear  implies  the  action  of  two  forces  in  parallel  planes 
and  very  near  together,  like  the  forces  in  a  pair  of  shears, 
from  which  analogy  the  name  is  derived.  Thus,  if  a 
bar  is  laid  upon  two  supports  and  two  loads,  each  P 
pounds,  are  applied  to  it  near  the  supports,  there  are  hence 
produced  near  each  support  two  parallel  forces  which 
tend  to  cut  the  bar  across  vertically  (Fig.  2).  In  each 
of  these  sections  the  shearing  stress  is  equal  to  P.  The 
deformation  caused  by  the  shearing  force  P  is  a  vertical 
sliding  between  the  upward  and  downward  forces;  and 


ART.  1  DIRECT   STRESSES  3 

the  bar  will  be  cut  across  if  the  external  shear  overcomes 
the  internal  stress. 

In  all  cases  of  direct  stress  the  total  stress  is  supposed, 
unless  otherwise  stated,  to  be  uniformly  distributed  over 


Fig.  2 

the  area  of  the  cross-section;  this  area  will  be  called  the 
'section  area.'  Thus  if  A  is  the  section  area  and  S  the 
unit-stress,  then 

P  =  AS,        S  =  j-,          A=^  (1) 

from  which  one  of  the  quantities  may  be  computed  when 
the  other  two  are  given.  For  example,  it  is  known  that  a 
wrought-iron  bar  will  rupture  under  tension  when  the 
unit-stress  S  becomes  50  000  pounds  per  square  inch  ; 
if  the  area  A  is  4l/2  square  inches,  then  the  tensile  force 
required  to  rupture  the  bar  is  P  =  4^X50  000  =  225  000 
pounds. 

Prob.  1  A.  A  cast-iron  bar  which  is  to  be  subjected  to  a  tension 
of  34  000  pounds  is  to  be  designed  so  that  the  unit-stress  shall  be 
2500  pounds  per  square  inch.  What  should  be  the  section  area  in 
square  inches?  If  the  bar  is  round,  what  should  be  its  diameter? 


I  Prob.  IB.     If  a  cast-iron  bar,  1MX2J^  inches  in  section  area, 

breaks  under  a  tension  of  66  000  pounds,  what  tension  will  probably 
break  a  bar  1J4  inches  in  diameter? 

Prob.  1  C.  What  should  be  the  size  of  a  round  bar  of  structural 
steel  to  carry  a  tension  of  200000  pounds  with  a  unit-stress  of 
15  000  pounds  per  square  inch? 


4  ELASTIC  AND  ULTIMATE  STRENGTH  Cfl.  1 

ART.  2.    THE  ELASTIC  LIMIT 

When  a  tensile  force  is  gradually  applied  to  a  bar  it 
elongates,  and  up  to  a  certain  limit  the  elongation  is 
proportional  to  the  force.  Thus,  if  a  bar  of  wrought  iron 
one  square  inch  in  section  area  and  100  inches  long  is 
subjected  to  a  tension  of  5000  pounds,  it  will  be  found  to 
elongate  0.02  inches;  if  10  000  pounds  is  applied,  the 
elongation  will  be  0.04  inches;  for  15  000  pounds  it  will 
be  0.06  inches,  for  20  000  pounds  0.08  inches,  for  25  000 
pounds  0.10  inches.  Thus  far  each  addition  of  5000 
pounds  has  produced  an  elongation  of  0.02  inches.  But 
when  the  next  5000  pounds  is  added,  making  a  total 
stress  of  30  000  pounds,  it  will  be  found  that  the  total 
elongation  is  about  0.13  or  0.14  inches,  and  hence  the 
elongations  are  increasing  more  rapidly  than  the  stresses. 

The  'elastic  limit'  is  defined  to  be  that  unit-stress  at 
which  the  deformations  begin  to  increase  in  a  faster  ratio 
than  the  stresses.  In  the  above  illustration  this  limit 
is  about  25  000  pounds  per  square  inch,  and  this  indeed 
is  the  average  value  of  the  elastic  limit  for  wrought  iron. 
The  term  'elastic  strength'  is  perhaps  more  expressive 
than  elastic  limit,  but  the  latter  is  the  one  in  general  use. 

When  the  unit-stress  in  a  bar  is  less  than  the  elastic 
limit  the  bar  returns,  when  the  stress  is  removed,  to  its 
original  length.  When  the  unit-stress  is  greater  than 
the  elastic  limit,  the  bar  does  not  fully  spring  back,  but 
there  remains  a  so-called  permanent  set.  In  other  words, 
the  elastic  properties  of  a  bar  are  injured  if  it  is  stressed 
beyond  the  elastic  limit.  Hence  it  is  a  fundamental  rule 
in  designing  engineering  constructions  that  the  unit- 


ART.  2 


THE    ELASTIC   LIMIT 


stresses  in  the  members  should  never  exceed  the  elastic 
limit   of   the   material. 

The  following  are  average  values  of  the  elastic  limits 
of  the  four  materials  most  used  in  engineering  construc- 
tion under  tensile  and  compressive  stresses. 
TABLE  1.     ELASTIC  LIMITS 


Material 

Pounds  per  Square  Inch 

Kilos  per  Sq.  Centimeter 

Tension 

Compres- 
sion 

Tension 

Compres- 
sion 

Timber 
Cast  Iron 
Wrought  Iron 
Steel,  structural 

3000 
6000 
25000 
35000 

3000 
20000 
25000 
35000 

210 
420 

1750 
2450 

210 
1400 
1750 
2450 

Values  in  English  units  must  be  carefully  kept  in  mind 
by  the  student;  and  it  may  be  noted  that  pounds  per 
square  inch  multiplied  by  0.07  will  give  kilos  per  square 
centimeter.  But  little  is  known  concerning  the  elastic 
limit  in  shear;  it  is  probably  about  three-fourths  of  the 
values  above  given  for  tension. 

Prob.  2  A.  A  steel  tie  rod  in  a  bridge  is  \Y±  inches  in  diameter. 
What  load  can  be  put  on  the  rod  if  the  unit-stress  is  not  to  exceed 
one-half  the  elastic  limit? 

Prob.  2  B.  A  square  stick  of  timber  is  to  carry  a  compressive 
load  of  81  000  pounds.  What  should  be  its  size  in  order  that  the 
unit-stress  may  be  one-third  of  the  elastic  limit? 

ART.  3.     ULTIMATE  STRENGTH 

When  a  bar  is  under  stress  exceeding  its  elastic  limit 
it  is  usually  in  an  unsafe  condition.  As  the  stress  is 
increased  by  the  application  of  exterior  forces  the  defor- 


6  ELASTIC   AND    ULTIMATE    STRENGTH  CH.  1. 

mation  rapidly  increases,  until  finally  the  rupture  of  the 
bar  occurs.  By  the  term  'ultimate  strength'  is  meant 
that  unit-stress  which  occurs  just  before  rupture,  it  being 
the  highest  unit-stress  that  the  bar  will  bear. 

The  ultimate  strengths  of  materials  are  from  two  to 
four  times  their  elastic  limits,  but  for  some  materials 
they  are  much  greater  in  compression  than  in  tension. 
The  average  values  of  the  ultimate  strengths  will  be  given 
in  subsequent  articles. 

The  'factor  of  safety'  is  a  number  which  results  by 
dividing  the  ultimate  strength  by  the  actual  unit-stress 
that  exists  in  a  bar.  For  example,  a  stick  of  timber, 
6X6  inches  in  section  area,  whose  ultimate  strength  in 
tension  is  10  000  pounds  per  square  inch,  is  under  a 
tensile  stress  of  32  400  pounds.  The  unit-stress  then  is 
32  400/36  =  900  pounds  per  square  inch,  and  the  factor 
of  safety  is  10000/900=11.  The  factor  of  safety  was 
formerly  much  used  in  designing,  but  it  is  now  considered 
the  better  plan  to  judge  of  the  security  of  a  body  under 
stress  by  reference  to  its  elastic  limit.  Thus  in  the  above 
case,  as  the  unit-stress  is  only  one-third  the  elastic  limit 
for  timber,  the  degree  of  security  may  be  regarded  as 
sufficient. 

Prob.  3  A.  A  bar  of  wrought  iron  2^  inches  in  diameter  ruptures 
under  a  tension  of  271  000  pounds.  What  is  its  ultimate  strength 
in  pounds  per  square  inch? 

Prob.  3  B.  What  should  be  the  size  of  a  round  bar  of  structural 
steel  to  carry  a  tension  of  125  000  pounds  with  a  factor  of  safety  of  5? 

Prob.  3  C.  What  force  is  required  to  rupture  in  tension  a  cast- 
iron  bar  8  inches  in  diameter,  the  ultimate  tensile  strength  of  cast 
iron  being  20  000  pounds  per  square  inch? 


ART.  4 


TENSION 


ART.  4.    TENSION 


When  a  bar  is  tested  under  tension,  it  is  done  by  loads 
which  are  gradually  applied.  The  elongations  increase 
proportionally  to  the  stresses  until  the  elastic  limit  is 
reached.  After  the  unit-stress  has  exceeded  the  elastic 
limit  the  elongations  increase  more  rapidly  than  the 
stresses,  and  a  reduction  in  area  of  the  cross-section  of 
the  bar  often  occurs.  Finally  the  ultimate  strength  of  the 
material  is  reached,  and  the  bar  tears  apart. 

A  graphical  illustration  of  these  phenomena  may  be 
made  by  laying  off  the  unit-stresses  as  ordinates  and 
the  elongations  per  unit  of  length  as  abscissas  (Fig.  3) .  At 


Fig.  3 

various  intervals,  as  the  test  progresses,  the  applied  loads 
are  observed  and  the  resulting  elongations  are  measured. 
The  loads  divided  by  the  section  area  give  the  unit- 
stresses,  while  the  total  elongations  divided  by  the  length 
of  the  bar  give  the  unit-elongations.  On  the  plot  a  point 


ELASTIC    AND    ULTIMATE    STRENGTH 


CH.I 


is  made  for  the  intersection  of  each  unit-stress  with  its 
corresponding  unit-elongation,  and  a  curve  is  drawn 
connecting  the  several  points  for  each  material.  In  this 
way  curves  are  plotted  showing  the  properties  of  each 
material.  It  is  seen  that  each  curve  is  a  straight  line 
from  the  origin  0  until  the  elastic  limit  is  reached,  showing 
that  the  elongations  increase  proportionally  to  the  unit- 
stresses.  At  the  elastic  limit  a  sudden  change  in  the  curve 
is  seen,  and  afterwards  the  elongation  increases  more 
rapidly  than  the  stress.  The  end  of  the  curve  indicates 
the  point  of  rupture.  The  curve  for  steel  in  the  diagram 
is  for  a  quality  much  stronger  than  structural  steel,  this 
being  the  kind  mostly  used  in  bridges  and  buildings. 

The  ultimate  elongation  is  an  index  of  the  ductility 
of  the  material,  and  is  hence  generally  recorded  for 
wrought  iron  and  steel;  this  is  usually  expressed  as  a 
percentage  of  the  total  length  of  the  bar,  or  it  is  100 
times  the  unit-elongation.  The  following  table  gives 
mean  values  of  the  ultimate  strengths  and  ultimate 
elongations  for  the  principal  materials  used  in  tension. 

TABLE  2.     TENSILE  STRENGTHS 


Material 

Ultimate  Strength 

Ultimate 
Elongation 
Per  Cent. 

Pounds  per 
Square  Inch 

Kilos  per 
Sq.  Centimeter 

Timber 
Cast  Iron 
Wrought  Iron 
Steel,  structural 

10000 

20000 
50000 
65000 

700 

1400 
3500 
4500 

1.5 
0.5 
30 
25 

All  these  values  should  be  regarded  as  rough  averages, 
since  they  are  subject  to  much  variation  with  different 


ART.  5  COMPRESSION  9 

qualities  of  the  material;  for  instance,  poor  timber  may 
be  as  low  as  6000,  while  strong  timber  may  be  as  high  as 
20  000  pounds  per  square  inch  in  ultimate  tensile  strength. 
The  ultimate  strengths  given  in  the  table  should,  how- 
ever, be  memorized  by  the  student  as  a  basis  for  future 
knowledge,  and  they  will  be  used  for  all  the  examples  and 
problems  in  this  book,  unless  otherwise  stated. 

Prob.  4  A.  What  should  be  the  diameter  of  a  wrought-iron  bar 
so  as  to  carry  a  tension  of  200  000  pounds  with  a  factor  of  safety 
of  5?  If  the  bar  is  cast  iron,  what  should  be  its  diameter?  V- 

Prob.  4  B.  A  bar  of  wrought  iron  one  square  inch  in  section  area 
and  one  yard  long  weighs  10  pounds.  Find  the  length  of  a  vertical 
bar  which  ruptures  under  its  own  weight  when  hung  at  its  upper  end. 

ART.  5.     COMPRESSION 

The  phenomena  of  compression  are  similar  to  those  of 
tension  provided  that  the  elastic  limit  is  not  exceeded, 
the  shortening  of  the  bar  being 
proportional    to    the    applied  |p 

force.  After  the  elastic  limit 
is  passed  the  shortening  in- 
creases more  rapidly  than  the 
stress.  When  the  length  of 
the  specimen  is  less  than  about 
ten  times  its  least  thickness, 
failure  usually  occurs  by  an 
oblique  splitting  or  shearing,  j-f 

as  seen  in  Fig.  4.    When  the  •Fig  4 

length  is  large  compared  with 

the  thickness,  failure  usually  occurs  under  a  sidewise 
bending,  so  that  this  is  not  a  case  of  simple  compression. 
All  the  values  given  in  the  following  table  refer  to  the 


10 


ELASTIC  AND    ULTIMATE   STRENGTH 


CH.I 


short  specimens;    longer  pieces  are  called  'columns'  or 
'struts,'  and  these  will  be  discussed  in  Chap.  5. 

The  mean  values  of  the  ultimate  compressive  strengths 
of  the  principal  materials  are  tabulated  below.  These  are 
subject  to  much  variation  in  different  qualities  of  the 
materials,  but  it  is  necessary  for  the  student  to  fix  them 
in  his  mind  as  a  preliminary  basis  for  more  extended 
knowledge.  It  is  seen  that  timber  is  not  quite  as  strong 
in  compression  as  in  tension,  that  cast  iron  is  ±V-2  tunes 
as  strong,  that  wrought  iron  and  structural  steel  have  the 
same  ultimate  strength  in  tension  and  compression. 
TABLE  3.  COMPRESSIVE  STRENGTHS 


Material 

Ultimate  Strength 

Pounds  per                   Kilos  per 
Square  Inch             Sq.  Centimeter 

Timber 

8000 

560 

Brick 

3000 

210 

Stone 

6000 

420 

Cast  Iron 

90000 

6300 

Wrought  Iron 
Steel,  structural 

50000 
65000 

3500 
4500 

The  investigation  of  a  body  under  compression  is 
made  by  formula  (1)  of  Art.  1.  For  example,  if  a  stone 
block  8X12  inches  in  cross-section  is  subjected  to  a 
compression  of  230  000  pounds,  the  unit-stress  produced 
is  230000/96  =  2400  pounds  per  square  inch,  and  the 
factor  of  safety  is  6000/2400  =  2%;  this  is  not  sufficiently 
high  for  stone,  as  will  be  seen  later. 

Prob.  5^4.  A  brick  2X4X8  inches  weighs  about  4J^  pounds. 
What  will  be  the  height  of  a  pile  of  bricks  so  that  the  unit-stress 
on  the  lowest  brick  shall  be  one-half  of  its  ultimate  strength? 


ART.  6 


SHEAR 


11 


Prob.  5  B.  A  short  cast-iron  column  is  12  inches  in  outside 
diameter  and  10  inches  in  inside  diameter.  Compute  its  factor  of 
safety  when  carrying  a  load  of  165  000  pounds. 

ART.  6.     SHEAR 

Shearing  stresses  exist  when  two  forces  acting  like  a 
pair  of  shears  tend  to  cut  a  body  between  them.  When 
a  hole  is  punched  in  a  plate,  the  ultimate  shearing  strength 
of  the  material  must  be  overcome.  If  two  thin  bars  are 
connected  by  a  rivet  and  then  are  subjected  to  tension, 
the  cross-section  of  the  rivet  between  the  plates  is  brought 
into  shear.  If  a  bolt  is  in  tension,  the  forces  acting  on  the 
head  tend  to  shear  or  strip  it  off. 

The  following  table  gives  the  average  ultimate  shearing 
strength  of  different  materials  as  determined  by  experi- 

TABLE  4.     SHEARING  STRENGTHS 


Material 

Ultimate  Strength 

Pounds  per                   Kilos  per 
Square  Inch            Sq.  Centimeter 

Timber/  longitudinal 
I  transverse 

600 

3000 

42 
210 

Cast  Iron 

20000 

1400 

Wrought  Iron 
Steel,  structural 

40000 
50000 

2800 
3500 

ment.  For  timber  this  is  much  smaller  along  the  grain 
than  across  the  grain;  in  the  first  direction  it  is  called 
the  longitudinal  shearing  strength,  and  in  the  second 
the  transverse  shearing  strength.  Rolled  plates  of 
wrought  iron  and  steel,  where  the  process  of  manufacture 
induces  a  fibrous  structure,  are  also  sheared  more  easily  in 
the  longitudinal  than  in  the  transverse  direction. 


12  ELASTIC  AND   ULTIMATE   STRENGTH  CH.  1 

Wooden  specimens  for  tensile  tests  like  that  shown 
in  Fig.  5  will  fail  by  shearing  off  the  ends  if  the  length  ab 
is  not  sufficiently  great.  For  instance,  suppose  ab  to  be 
6  inches,  and  the  diameter  of  the  central  part  to  be  2  inches. 


Fig.  5 

The  ends  are  grasped  tightly  by  the  machine  and  the 
cross-section  of  the  central  part  thus  brought  under 
tensile  stress.  The  force  required  to  cause  rupture  by 
tension  is 

P  =  vl£  =  3.14Xl2AXlOOOO  =  31  400  pounds. 

But  the  ends  also  tend  to  shear  off  along  the  surface  of 
a  cylinder  whose  diameter  is  2  inches  and  whose  length 
is  ab;  the  force  required  to  cause  rupture  by  shearing  on 
this  surface  is 

P  =  ,4/S  =  3.14X2X6XGOO  =  22  600  pounds. 

and  hence  the  specimen  will  fail  by  shearing  off  the  ends. 
To  prevent  this  the  distance  ab  must  be  made  longer 
than  6  inches. 

Prob.  6  A.  The  beam  in  Fig.  2  is  3  X4  inches  in  section-area,  and 
P  is  13  000  pounds.  Compute  the  shearing  unit-stress. 

Prob.  6  B.  A  wrought-iron  bolt  1 J^  inches  in  diameter  has  a  head 
1 J4  inches  long.  When  a  tension  of  15  000  pounds  is  applied  to  the 
bolt,  find  the  tensile  unit-stress  and  the  factor  of  safety  for  tension. 
Also  find  the  unit-stress  tending  to  shear  off  the  head  of  the  bolt, 
and  the  factor  of  safety  against  shear. 

ART.  7.    WORKING  UNIT-STRESSES 

When  a  body  of  cross-section  A  is  under  a  stress  P, 
the  unit-stress  S  produced  is  found  by  dividing  P  by 


ART.  7  WORKING  UNIT-STRESSES  13 

A.  By  comparing  this  value  of  S  with  the  ultimate 
strengths  and  elastic  limits  given  in  the  preceding  articles, 
the  degree  of  security  may  be  inferred.  This  process  is 
called  investigation.  The  student  may  not  at  first  be  able 
to  form  a  good  judgment  with  regard  to  the  degree  of 
security,  this  being  a  matter  which  involves  some  expe- 
rience as  well  as  acquaintance  with  engineering  precedents 
and  practice.  As  his  knowledge  increases,  however,  his 
ability  to  judge  whether  unit-stresses  are  or  are  not  too 
great  will  constantly  improve. 

When  a  body  is  to  be  designed  to  stand  a  total  stress  P, 
the  unit-stress  S  is  first  assumed  in  accordance  with  the 
rules  of  practice,  and  then  the  section  area  A  is  computed. 
Such  assumed  unit-stresses  are  often  called  working 
unit-stresses,  meaning  that  these  are  the  unit-stresses 
under  which  the  material  is  to  act  or  work.  In  selecting 
them,  two  fundamental  rules  are  to  be  kept  in  mind : 

1.  They  should  be  considerably  less  than  the  elastic  limits. 

2.  They  should  be  smaller  for  sudden  stresses  than  for  steady 

stresses. 

The  reason  for  the  first  requirement  is  given  in  Art.  2. 
The  reason  for  the  second  requirement  is  £hat  experience 
teaches  that  suddenly-applied  loads  and  shocks  are  more 
injurious  and  produce  higher  unit-stresses  than  steady 
loads.  Thus  a  bridge  subject  to  the  traffic  of  heavy 
trains  must  be  designed  with  lower  unit-stresses  than  a 
roof  where  the  variable  load  consists  only  of  snow  and 
wind. 

It  will  be  best  for  the  student  to  begin  to  form  his 
engineering  judgment  by  fixing  in  mind  the  following 
average  values  of  the  factors  of  safety  to  be  used  for 


14 


ELASTIC   AND    ULTIMATE    STRENGTH 


CH.  1 


different  materials  under  different  circumstances.     The 
working  unit-stress  will  then  be  found  for  any  special 

TABLE  5.     FACTORS  OF  SAFETY 


Material 

For  Steady 
Stress 
(Buildings) 

For  Varying        For  ghocka 
(Bridges)            (Machines) 

Timber 

8 

10 

15 

Brick  and  Stone 

15 

25 

35 

Cast  Iron 

6 

15 

20 

Wrought  Iron 
Steel,  structural 

4 
4 

6 
6 

10 
10 

case  by  dividing  the  ultimate  strengths  by  these  factors 
of  safety.  For  instance,  a  short  timber  strut  in  a  bridge 
should  have  a  working  unit-stress  of  about  8000/10  =  800 
pounds  per  square  inch. 

It  frequently  happens  that  a  designer  works  under 
specifications  in  which  the  unit-stresses  to  be  used  are 
definitely  stated.  The  writer  of  the  specifications  must 
necessarily  be  an  engineer  of  much  experience  and  with  a 
thorough  knowledge  of  the  best  practice.  It  may  be 
noted  also  that  the  particular  qualities  of  timber  or  steel 
to  be  used  will  influence  the  selection  of  working  unit- 
stresses,  and  in  fact  different  members  of  a  bridge  truss 
are  often  designed  with  different  unit-stresses.  The  two 
fundamental  rules  above  stated  are,  however,  the  guiding 
ones  in  all  cases. 

Modern  engineering  is  the  art  of  economic  construction. 
In  numerous  instances  this  will  be  secured  by  making 
all  parts  of  a  structure  of  equal  strength,  for  if  one  part 
is  stronger  than  another  it  has  an  excess  of  material 
which  might  have  been  saved. 


ART.  8  REVIEW  PROBLEMS  15 

Prob.  7  A.  A  wrought-iron  rod  is  to  be  under  a  stress  of  82  000 
pounds.  Find  its  diameter  when  it  is  to  be  used  in  a  building^  and 
also  when  it  is  to  be  used  in  a  bridge. 

Prob.  7  B.  The  total  shear  on  each  rivet  of  a  lap-riveted  joint 
is  2000  pounds.  If  the  rivet  is  Y%  inches  in  diameter,  find  the  factor 
of  safety  against  shearing. 

ART.  8.     REVIEW  PROBLEMS 

The  following  problems  may  serve  to  test  the  student 
as  to  his  knowledge  of  the  preceding  principles  and 
methods.  In  solving  problems  it  is  very  desirable  that 
a  neat  and  systematic  method  should  be  followed.  The 
practice  of  making  computations  with  a  pencil  on  loose 
scraps  of  paper  should  be  discontinued  by  every  student 
who  has  followed  it,  and  he  should  hereafter  solve  his 
problems  in  a  special  book,  using  pen  and  ink.  Before 
beginning  the  solution,  a  diagram  should  be  drawn  when- 
ever possible,  for  a  diagram  helps  the  student  to  under- 
stand the  problem,  and  a  problem  thoroughly  understood 
is  really  half  solved.  Before  beginning  the  computation 
of  a  numerical  problem,  it  is  best  to  make  a  mental 
estimate  of  the  answer,  for  thus  the  engineering  judge- 
ment of  the  student  will  be  developed.  In  Art.  54  will 
be  found  a  few  answers,  but  the  student  should  never 
look  there  for  an  answer  to  a  problem  until  he  has  com- 
pleted its  solution. 

Prob.  8  A.  During  the  tensile  test  of  a  steel  bar  %  inches  in 
diameter  the  load  at  the  elastic  limit  was  found  to  be  17  600  pounds. 
What  was  the  elastic  limit  in  pounds  per  square  inch? 

Prob.  8  B.  The  pull  on  the  piston-rod  of  a  steam  engine  is  25  000 
pounds.  If  the  diameter  of  the  rod  is  2J4  inches,  compute  the 
factor  of  safety. 

Prob.  8  C. '  During  the  tensile  test  of  a  cast-iron  bar  1  inch  in 


16  ELASTIC    AND    ULTIMATE    STRENGTH  CH.   1 

diameter  rupture  occurred  under  a  load  of  18  000  pounds.    What 
was  the  tensile  strength  of  the  cast  iron? 

Prob.  8  D.  In  a  test  on  a  wooden  specimen  of  shape  shown  in 
Fig.  5  a  load  of  1000  pounds  was  placed.  Diameter  of  specimen 
was  l^i  inches,  and  length  of  head  was  6  inches.  Find  the  tensile 
unit-stress  and  the  shearing  unit-stress. 

Prob.  8  E.  The  piston  of  an  engine  is  12  inches  in  diameter,  and 
the  diameter  of  the  piston  rod  is  2J4  inches.  The  maximum  steam 
pressure  is  120  pounds  per  sq.  in.  Find  the  tensile  unit-stress  on 
the  rod  and  the  factor  of  safety. 

Prob.  8  F.  A  wrought-iron  bolt  1^  inches  in  diameter  has  a  head 
1  inch  long.  Find  the  unit-stress  tending  to  shear  off  the  head  when 
a  tension  of  3000  pounds  is  applied  to  the  bolt. 

Prob.  8  G.  A  pipe-rack  in  a  shop  is  supported  by  four  wrought- 
iron  rods,  each  1  inch  in  diameter.  The  total  load  supported  by 
the  rods  is  two  long  tons.  Is  the  structure  safe? 

Prob.  8  H.  A  steel  column  is  supported  by  a  stone  base.  If  the 
total  load  on  the  base  is  30  000  pounds,  and  the  factor  of  safety 
is  to  be  10,  find  the  cross-section  area  of  the  stone. 

Prob.  8  /.  A  steel  tie  rod  in  a  roof  truss  is  to  sustain  a  load  of 
6000  pounds.  Find  the  size  of  rod  so  that  it  shall  have  a  factor  of 
safety  of  8. 

Prob.  8  K.  The  total  load  on  a  vertical  shaft  hanger  is  2000 
pounds.  The  hanger  is  held  in  place  by  means  of  two  1-inch  bolts. 
The  length  of  head  of  each  bolt  is  1^  inches.  Find  the  factor  of 
safety  against  shear  and  tension. 

Prob.  8  L.  A  bridge  carrying  a  total  load  of  320  000  pounds 
rests  on  two  stone  abutments.  The  bridge  rests  on  four  cast-iron 
bearing  plates.  Find  the  size  of  these  plates  so  that  the  stone  shall 
have  a  factor  of  safety  of  ten. 


ART.  9 


AVERAGE   WEIGHTS 


17 


CHAPTER  II 

GENERAL  PROPERTIES 

ART.  9.    AVERAGE  WEIGHTS 

The  average  weights  of  the  six  principal  materials 
used  in  engineering  constructions  are  given  in  the  following 
table,  together  with  their  specific  gravities.  These  are 
subject  to  more  or  less  variation,  according  to  the  quality 
of  the  material.  For  instance,  brick  may  weigh  as  low 

TABLE  6.     WEIGHT 


Material 

Pounds  per 
Cubic  Foot 

Kilos  per 
Cubic  Meter 

Specific 
Gravity 

Timber 

40 

600 

0.6 

Brick 

125 

2000 

2.0 

Stone 

160 

2600 

2.6 

Cast  Iron 

450 

7200 

7.2 

Wrought  Iron 

480 

7700 

7.7 

Steel 

490 

7800 

7.8 

as  100  or  as  high  as  150  pounds  per  cubic  foot,  depending 
upon  whether  it  is  soft  common  quality  or  fine  hard- 
pressed.  Unless  otherwise  stated,  the  above  values  will 
be  used  in  all  the  examples  and  problems  in  the  following 
pages,  and  hence  those  in  pounds  per  cubic  foot  must  be 
carefully  kept  in  the  memory. 

For  computing  the  weights  of  bars,  beams,  and  pieces 
of  uniform  section-area,  the  following  approximate  simple 
rules  are  much  used  by  engineers: 


18  GENERAL   PROPERTIES  CH.  2 

A  wrought-iron  bar  one  square  inch  in  section-area  and  one 

yard  long  weighs  ten  pounds. 
Timber  is  one-twelfth  the  weight  of  wrought  iron. 
Brick  is  one-fourth  the  weight  of  wrought  iron. 
Stone  is  one-third  the  weight  of  wrought  iron. 
Cast  iron  is  six  percent  lighter  than  wrought  iron. 
Steel  is  two  percent  heavier  than  wrought  iron. 

For  example,  if  a  bar  of  wrought  iron  be  1^X3  inches 
in  section  and  22  feet  long,  its  section-area  is  4%  square 
inches  and  its  weight  is  45X7%  =  330  pounds.  A  steel 
bar  of  the  same  dimensions  will  weigh  330+0.02X330  = 
337  pounds,  and  a  cast-iron  bar  will  weigh  330  — 0.06  X 
330  =  3 10  pounds. 

By  reversing  the  above  rules  the  section-areas  are 
readily  found  when  the  weights  per  linear  yard  are  given. 
Thus,  if  a  stick  of  timber  15  feet  long  weighs  120  pounds, 
its  weight  per  yard  is  24  pounds  and  its  section-area  is 
2.4X12  =  28.8  square  inches. 

Prob.  9  A.  What  is  the  weight  of  a  stone  block  12X18  inches 
and  4:^2  feet  long?  How  many  square  inches  in  the  cross-section 
of  a  steel  railroad  rail  which  weighs  95  pounds  per  yard? 

Prob.  9  B.  If  a  cast-iron  water  pipe  12  feet  long  weighs  1000 
pounds,  what  is  its  section-area?  Find  the  diameter  of  a  wrought- 
iron  bar  which  is  24  feet  long  and  weighs  1344  pounds. 

ART.  10.    TESTING  MACHINES 

The  simplest  method  of  testing  is  by  tension,  a  speci- 
men being  used  like  that  shown  in  Art.  6.  The  heads 
are  either  gripped  in  jaws,  or  they  are  provided  with 
threads  so  that  they  may  be  screwed  into  nuts  to  which 
the  forces  are  applied.  The  power  may  be  furnished 
by  a  lever,  a  screw,  or  by  hydraulic  pressure,  the  last 


ART.  10  TESTING   MACHINES  19 

being  the  method  in  machines  of  high  capacity.  In  these 
tests  the  elastic  limit,  ultimate  strength,  and  the  ultimate 
elongation  are  generally  recorded,  the  latter  being  ex- 
pressed as  a  percentage  of  the  original  length.  For  ductile 
materials  the  contraction  of  area  of  the  fractured  speci- 
men is  also  noted,  as  this  does  not  vary  with  the  length 
of  the  specimen  to  the  same  extent  as  the  ultimate  elonga- 
tion. In  such  tensile  tests  the  load  is  applied  gradually, 
and  not  suddenly  or  with  impact. 

The  elastic  limit  is  detected  by  taking  a  number  of 
measurements  of  the  elongation  for  different  loads,  and 
then  noting  when  these  begin  to  vary  more  rapidly  than 
the  stresses.  For  ductile  materials  the  change  is  a  sudden 
one,  and  it  may  be  often  noted  by  the  drop  of  the  scale 
beam  of  the  machine. 

Compressive  tests  are  confined  mainly  to  brick  and 
stone,  and  are  but  little  used  for  commercial  tests  of 
metals  on  account  of  the  difficulty  of  securing  a  uniform 
distribution  of  pressure  over  the  surfaces.  Cement, 
which  is  always  used  in  compression,  is  indeed  usually 
tested  by  tension,  this  being  found  to  be  the  cheaper  and 
more  satisfactory  method. 

The  capacity  of  a  testing  machine  is  the  number  of 
pounds  it  can  exert  as  tension  or  compression.  A  small 
machine  for  testing  wire  or  cement  need  not  have  a 
capacity  greater  than  1000  or  2000  pounds.  Machines 
of  50  000,  100  000,  and  150  000  pounds  for  testing  metals 
are  common.  The  Watertown  machine  has  a  capacity  of 
800  000  pounds,  and  can  test  a  small  hair  or  a  steel  bar 
of  10  square  inches  section-area  with  equal  precision. 
A  list  of  the  fourteen  largest  testing  machines  in  the 


20 


GENERAL   PROPERTIES 


CH.  2 


United    States  is  given  in    American    Civil    Engineers' 
Pocket  Book  (New  York,  1912). 

Fig.  6  shows  an  Olsen  screw  testing  machine  of  40  000 
pounds  capacity.  The  power  is  applied  by  hand  by  means 
of  the  crank  on  the  left,  and  this  causes  the  four  vertical 


Fig.  6 

screws  to  have  a  slow  upward  or  downward  movement. 
The  upper  ends  of  the  screws  are  fastened  to  a  table  A, 
which  hence  partakes  of  the  vertical  motion.  When  a 
tensile  test  is  to  be  made,  one  end  of  the  specimen  is 
gripped  by  jaws  in  the  movable  table  A,  and  the  other 
end  by  jaws  in  the  upper  fixed  table  B;  in  the  figure 
a  tensile  specimen  is  seen  in  this  position.  The  crank  is 
then  turned  so  as  to  cause  the  vertical  screws  and  the 
movable  table  to  descend,  and  thus  a  stress  is  brought 


ART.  11  TIMBER  21 

upon  the  specimen.  For  a  compressive  test  the  specimen 
is  placed  between  the  lower  fixed  table  C  and  the  movable 
table  A,  the  latter  being  caused  to  descend  by  turning 
the  crank  as  before.  The  load  applied  to  the  specimen 
at  any  instant  is  weighed  on  the  lever  scale  at  the  right 
by  moving  the  weight  D  so  that  the  scale  arm  will  balance. 
Machines  of  greater  capacity  than  40000  pounds  are 
usually  operated  by  power,  which  is  transmitted  from 
a  motor  to  a  pulley  on  the  shaft  of  the  machine. 

Tests  are  also  made  by  loading  beams  transversely  and 
measuring  the  deflections,  as  well  as  finding  the  load 
required  to  produce  rupture.  The  machine  shown  in 
Fig.  6  can  be  used  for  flexural  tests  of  short  beams  by 
placing  two  supports  on  the  lower  fixed  table,  while  for 
long  beams  a  special  attachment  can  be  made  to  this 
table.  In  both  cases  the  load  is  applied  by  lowering  the 
movable  table. 

Prob.  10  A.  What  is  the  diameter  of  the  largest  bar  of  structural 
steel  which  can  be  tested  in  a  machine  of  100  000  pounds  capacity? 

Prob.  10  B.  A  steel  eye-bar  tested  at  Phcenixville  was  10X2^ 
inches  in  size  and  47  feet  long.  The  length  after  rupture  was  57.6 
feet,  and  the  area  of  the  fractured  cross-section  was  13.0  square 
inches.  Compute  the  percentage  of  ultimate  elongation  and  the 
percentage  of  reduction  of  area. 

ART.  11.    TIMBER 

Good  timber  is  of  uniform  color  and  texture,  free  from 
knots,  sap  wood,  wind  shakes,  and  decay.  It  should  be 
well  seasoned,  which  is  best  done  by  exposing  it  to  the 
sun  and  wind  for  two  or  three  years  to  dry  out  the  sap. 
The  heaviest  timber  is  usually  the  strongest;  also  the 
darker  the  color  and  the  closer  the  annular  rings,  the 


22 


GENERAL   PROPERTIES 


CH.  2 


stronger  and  better  it  is,  other  things  being  equal.  The 
strength  of  timber  is  always  greatest  in  the  direction  of 
the  grain,  the  sidewise  resistance  to  tension  or  compression 
being  scarcely  one-fourth  of  the  longitudinal. 

The  following  table  which  gives  average  values  of  the 
ultimate  strength  of  a  few  of  the  common  kinds  of  timber 
will  be  useful  for  reference.  These  values  have  been 
determined  from  tests  of  small  specimens  carefully 
selected  and  dried.  Large  pieces  of  timber  such  as  are 

TABLE  7.     STRENGTH  OF  TIMBER 


Kind 

Pounds 
per  Cubic 
Foot 

Pounds  per  Square  Inch 

Tensile 
Strength 

Compres- 
sive 
Strength 

Hemlock 
White  Pine 
Chestnut 
Red  Oak 
Yellow  Pine 
White  Oak 

25 

27 
40 
42 
45 

48 

8000 
8000 
12000 
9000 
15000 
12000 

5000 
5500 
5000 
6000 
9000 
8000 

actually  used  in  engineering  structures  will  probably 
have  an  ultimate  strength  of  from  fifty  to  eighty  per  cent 
of  these  values.  Moreover,  the  figures  are  liable  to  a 
range  of  25  per  cent  on  account  of  variations  in  quality 
and  condition  arising  from  place  of  growth,  time  when  cut, 
and  method  of  seasoning.  To  cover  these  variations  the 
factor  of  safety  of  10  is  not  too  high,  even  for  steady 
stresses. 

The  shearing  strength  of  timber  is  still  more  variable 
than  the  tensile  or  compressive  resistance.  White  pine 
across  the  grain  may  be  put  at  2500  pounds  per  square 
inch,  and  along  the  grain  at  500.  Chestnut  has  1500 


ART.  12  BRICK  23 

and  600  respectively,  yellow  pine  and  oak  perhaps  4000 
and  600  respectively. 

The  elastic  limit  of  timber  is  poorly  defined.  In  precise 
tests  on  good  specimens  it  is  sometimes  observed  at 
about  one-half  the  ultimate  strength,  but  under  ordinary 
conditions  it  is  safer  to  put  it  at  one-third.  The  ultimate 
elongation  is  small,  usually  being  between  1  and  2  per 
cent. 

Prob.  11  A.  What  should  be  the  size  of  a  short  piece  of  yellow 
pine  which  is  to  carry  a  steady  load  of  80  000  pounds? 

Prob.  11  B.  If  a  piece  of  white  cedar  2X2  inches  in  cross-section 
ruptures  under  a  compression  of  20  800  pounds,  what  is  the  size  of 
a  square  section  that  will  stand  25  000  pounds  with  a  factor  of 
safety  of  10  ? 

ART.  12.     BRICK 

Brick  is  made  of  clay  which  consists  mainly  of  silicate 
of  alumina  with  compounds  of  lime,  magnesia,  and  iron. 
The  clay  is  prepared  by  cleaning  it  carefully  from  pebbles 
and  sand,  mixing  it  with  about  one-half  its  volume  of 
water,  and  tempering  it  by  hand  stirring  or  in  a  pug  mill. 
It  is  then  moulded  in  rectangular  boxes  by  hand  or  by 
special  machines,  and  the  green  bricks  are  placed  under 
open  sheds  to  dry.  These  are  piled  in  a  kiln  and  heated 
for  nearly  two  weeks  until  those  nearest  to  the  fuel 
assume  a  partially  vitrified  appearance. 

Three  qualities  of  brick  are  taken  from  the  kiln;  'arch- 
brick'  are  those  from  around  the  arches  where  the  fuel 
is  burned,  these  are  hard  and  often  brittle;  'body-brick,' 
from  the  interior  of  the  kiln,  are  of  the  best  quality; 
'soft  brick,'  from  the  exterior  of  the  pile,  are  weak  and 
only  suitable  for  filling.  Paving  brick  are  burned  in 


24  GENERAL   PROPERTIES  CH.   2 

special  kilns,  often  by  natural  gas  or  by  oil,  the  rate  of 
heating  being  such  as  to  insure  toughness  and  hardness. 

The  common  size  is  2X4X81/i  inches,  and  the  average 
weight  4%  pounds.  A  pressed  brick,  however,  may  weigh 
nearly  5%  pounds.  Good  bricks  should  be  of  regular 
shape,  have  parallel  and  plane  faces,  with  sharp  angles 
and  edges.  They  should  be  of  uniform  texture,  and  when 
struck  a  quick  blow  should  give  a  sharp,  metallic  ring. 
The  heavier  the  brick,  other  things  being  equal,  the 
stronger  and  better  it  is. 

Poor  brick  will  absorb  when  dry  from  20  to  30  per 
cent  of  its  weight  of  water,  ordinary  qualities  absorb 
from  10  to  20  per  cent,  while  hard  paving  brick  should 
not  absorb  more  than  2  or  3  per  cent.  An  absorption 
test  is  valuable  in  measuring  the  capacity  of  brick  to 
resist  the  disintegrating  action  of  frost,  and  as  a  rough 
general  rule,  the  greater  the  amount  of  water  absorbed 
the  less  is  the  strength  and  durability. 

The  crushing  strength  of  brick  is  variable;  while  a 
mean  value  may  be  3000  pounds  per  square  inch,  soft 
brick  will  scarcely  stand  500,  pressed  brick  may  run  to 
10  000,  and  the  best  qualities  of  paving  brick  have  given 
15  000  pounds  per  square  inch,  or  even  more.  Crushing 
tests  are  usually  made  on  whole  or  half-bricks  and  are 
hence  lacking  in  precision,  since  opposite  surfaces  are 
rarely  truly  parallel.  Tensile  and  shearing  tests  of 
bricks  are  rarely  made,  and  but  little  is  known  of  their 
behavior  under  such  stresses;  the  ultimate  tensile  strength 
may  perhaps  range  from  50  to  500  pounds  per  square  inch. 

Prob.  12  A.  Compute  the  unit-stress  at  the  base  of  a  brick  wall 
17  inches  thick  and  55  feet  high.  What  is  the  factor  of  safety? 


ART.  13  STONE  25 

Prob.  12  B.  A  brick  weighs  4.42  pounds  when  dry  and  4.75 
pounds  after  immersion  for  one  day  in  water.  What  percentage  of 
water  has  it  absorbed?  «1  d  ^ 

ART.  13.     STONE 

Sandstone,  as  its  name  implies,  is  sand,  usually  quartzite, 
which  has  been  consolidated  under  heat  and  pressure. 
It  varies  much  in  color,  strength,  and  durability,  but 
many  varieties  form  most  valuable  building  material. 
In  general  it  is  easy  to  cut  and  dress,  but  the  variety 
known  as  Potsdam  sandstone  is  very  hard  in  some 
localities. 

Limestone  is  formed  by  consolidated  marine  shells, 
and  is  of  diverse  quality.  Marble  is  limestone  which 
has  been  reworked  by  the  forces  of  nature  so  as  to  expel 
the  impurities,  and  leave  a  nearly  pure  carbonate  of  lime; 
it  takes  a  high  polish,  is  easily  cut,  and  makes  one  of  the 
most  beautiful  building  stones. 

Granite  is  a  rock  of  aqueous  origin  metamorphosed 
under  heat  and  pressure;  its  composition  is  quartz,  feld- 
spar, and  mica,  but  in  the  variety  called  gneiss  the  mica 
is  replaced  by  hornblende.  It  is  fairly  easy  to  work, 
usually  strong  and  durable,  and  some  varieties  will  take 
a  high  polish. 

Trap,  or  basalt,  is  'an  igneous  rock  without  cleavage. 
It  is  hard  and  tough,  and  less  suitable  for  building  con- 
structions than  other  rocks,  as  large  blocks  cannot  be 
readily  obtained  and  cut  to  size.  It  has,  however,  a  high 
strength,  and  is  remarkable  for  durability. 

The  average  weight  of  sandstone  is  about  150,  of 
limestone  160,  of  granite  165,  and  of  trap  175  pounds 
per  cubic  foot.  The  ultimate  compressive  strength  of 


26  GENERAL   PROPERTIES  Cfl.  2 

sandstone  is  about  5000,  of  limestone  7000,  of  granite 
12  000,  and  of  trap  16  000  pounds  per  square  inch;  these 
figures  refer  to  small  blocks,  but  the  ultimate  strength  of 
large  blocks  is  materially  smaller. 

The  quality  of  a  building  stone  cannot  be  safely  inferred 
from  tests  of  strength,  as  its  durability  depends  largely 
upon  its  capacity  to  resist  the  action  of  the  weather. 
Hence  corrosion  and  freezing  tests,  impact  tests,  and 
observations  of  the  behavior  of  stone  under  conditions 
of  actual  use  are  more  important  than  the  determination 
of  crushing  strength  in  a  compression  machine. 

Prob.  13  A.  Find  the  weight  of  a  granite  column  18  feet  high 
and  18  inches  in  diameter. 

Prob.  13  B.  A  stone  pier  12X30  feet  at  the  base,  8X24  feet  at 
the  top,  and  16^  feet  high  is  to  be  built  at  $6.37  per  cubic  yard. 
What  is  the  total  cost? 

ART.  14.     CAST  IRON 

Cast  iron  is  a  modern  product,  having  been  first  made 
in  England  about  the  beginning  of  the  fifteenth  century. 
Ores  of  iron  are  melted  in  a  blast  furnace,  producing 
pig  iron.  The  pig  iron  is  remelted  in  a  cupola  furnace 
and  poured  into  moulds,  thus  forming  castings.  Beams, 
columns,  pipes,  braces,  and  blocks  of  every  shape  required 
in  engineering  structures  are  thus  produced. 

Pig  iron  is  divided  into  two  classes,  foundry  pig  and 
forge  pig,  the  former  being  used  for  castings  and  the  latter 
for  making  wrought  iron.  Foundry  pig  has  a  dark-gray 
fracture,  with  large  crystals  and  a  metallic  luster;  forge 
pig  has  a  light-gray  or  silver-white  fracture,  with  small 
crystals.  Foundry  pig  has  a  specific  gravity  of  from  7.1 


ART.  14  CAST  IRON  27 

to  7.2,  and  it  contains  from  6  to  4  per  cent  of  carbon; 
forge  pig  has  a  specific  gravity  of  from  7.2  to  7.4,  and  it 
contains  from  4  to  2  per  cent  of  carbon.  The  higher  the 
percentage  of  carbon  the  less  is  the  specific  gravity,  and 
the  easier  it  is  to  melt  the  pig.  Besides  the  carbon  there 
are  present  from  1  to  5  per  cent  of  other  impurities,  such 
as  silicon,  manganese,  and  phosphorus. 

The  properties  and  strength  of  castings  depend  upon 
the  quality  of  the  ores  and  the  method  of  their  manu- 
facture in  both  the  blast  and  the  cupola  furnace.  Cold 
blast  pig  produces  stronger  iron  than  the  hot  blast,  but 
it  is  more  expensive.  Long  continued  fusion  improves  the 
quality  of  the  product,  as  also  do  repeated  meltings. 
The  darkest  grades  of  foundry  pig  make  the  smoothest 
castings,  but  they  are  apt  to  be  brittle;  the  light-gray 
grades  make  tough  castings,  but  they  are  apt  to  contain 
blow  holes  or  imperfections. 

The  percentage  of  carbon  in  cast  iron  is  a  controlling 
factor  which  governs  its  strength,  particularly  that 
percentage  which  is  chemically  combined  with  the  iron. 
As  average  values  for  the  ultimate  strength  of  cast  iron, 
20  000  and  90  000  pounds  per  square  inch  in  tension  and 
compression  respectively  are  good  figures.  In  any  par- 
ticular case,  however,  a  variation  of  from  10  to  20  per 
cent  from  these  values  may  be  expected,  owing  to  the  great 
variation  in  quality.  The  elastic  limit  is  poorly  defined, 
there  being  no  sudden  increase  in  deformation,  as  in 
ductile  materials. 

The  high  compressive  strength  and  cheapness  of  cast 
iron  render  it  a  valuable  material  for  many  purposes; 
but  its  brittleness,  low  tensile  strength,  and  low  ductility 


28  GENERAL  PROPERTIES  CH.  2 

forbid  its  use  in  structures  subject  to  variations  of  load 
or  to  shocks.  Its  ultimate  elongation  being  scarcely 
one  per  cent,  the  work  required  to  cause  rupture  in 
tension  is  small  compared  to  that  for  wrought  iron  and 
steel,  and  hence  as  a  structural  material  the  use  of  cast 
iron  must  be  confined  entirely  to  cases  of  compression. 

Prob.  14  A.  A  cast-iron  bar  weighing  31  pounds  per  linear  yard 
is  to  be  subjected  to  tension.  How  many  pounds  are  required  to 
rupture  it? 

Prob.  14  B.  What  must  be  the  capacity  of  a  testing  machine 
to  break  a  cast-iron  block  2  inches  square? 

ART.  15.     WROUGHT  IRON 

The  ancient  peoples  of  Europe  and  Asia  were  acquainted 
with  wrought  iron  and  steel  to  a  limited  extent.  It  is 
mentioned  in  Genesis,  iv,  22,  and  in  one  of  the  oldest 
pyramids  of  Egypt  a  piece  of  iron  has  been  found.  It 
was  produced,  probably,  by  the  action  of  a  hot  fire  on 
very  pure  ore.  The  ancient  Britons  built  bloomaries 
on  the  tops  of  high  hills,  a  tunnel  opening  toward  the 
north  furnishing  a  draft  for  the  fire,  which  caused  the 
carbon  and  other  impurities  to  be  expelled  from  the  ore, 
leaving  behind  nearly  pure  metallic  iron. 

Modern  methods  of  manufacturing  wrought  iron  are 
mainly  by  the  use  of  forge  pig  (Art.  14),  the  one  most 
extensively  used  being  the  puddling  process.  Here  the 
forge  pig  is  subjected  to  the  oxidizing  flame  of  a  blast 
in  a  reverberatory  furnace,  where  it  is  formed  into  pasty 
balls  by  the  puddler.  A  ball  taken  from  the  furnace 
is  run  through  a  squeezer  to  expel  the  cinder  and  then 
rolled  into  a  muck  bar.  The  muck  bars  are  cut,  laid  in 


ART.  15  WROUGHT  IRON  29 

piles,  heated,  and  rolled,  forming  what  is  called  merchant 
bar.  If  this  is  cut,  piled,  and  rolled  again,  a  better  prod- 
uct, called  best  iron,  is  produced.  A  third  rolling  gives 
'best  best'  iron,  a  superior  quality,  but  high  in  price. 

Tlie  product  of  the  rolling-mill  is  bar  iron,  plate  iron, 
shape  iron,  beams,  and  rails.  Bar  iron  is  round,  square, 
and  rectangular  in  section;  plate  iron  is  from  J4  to  1  inch 
thick,  and  of  varying  widths  and  lengths;  shape  iron 
includes  angles,  tees,  channels,  and  other  forms  used  in 
structural  work;  beams  are  I-shaped,  and  of  the  deck  or 
rail  form.  Structural  shapes  and  beams  are,  however, 
now  almost  entirely  rolled  in  mild  steel. 

Wrought  iron  is  metallic  iron  containing  less  than  0.25 
per  cent  of  carbon,  and  which  has  been  manufactured 
without  fusion.  Its  tensile  and  compressive  strengths 
are  closely  equal,  and  range  from  50  000  to  60  000  pounds 
per  square  inch.  The  elastic  limit  is  well  defined  at  about 
25  000  pounds  per  square  inch,  and  within  that  limit  the 
law  of  proportionality  of  stress  to  deformation  is  strictly 
observed.  It  is  tough  and  ductile,  having  an  ultimate 
elongation  of  from  20  to  30  per  cent.  It  is  malleable, 
can  be  forged  and  welded,  and  has  a  high  capacity  to 
withstand  the  action  of  shocks.  It  cannot,  however,  be 
tempered  so  as  to  be  used  for  cutting  tools. 

The  cold-bend  test  for  wrought  iron  is  an  important 
one  for  judging  of  general  quality.  A  bar  perhaps  3/iX% 
inches  and  15  inches  long  is  bent  when  cold  either  by 
pressure  or  by  blows  of  a  hammer.  Bridge  iron  should 
bend  through  an  angle  of  90  degrees  to  a  curve  whose 
radius  is  twice  the  thickness  of  the  bar,  without  cracking. 
Rivet  iron  should  bend  through  180  degrees  until  the 


30  GENERAL   PROPERTIES  CH.  2 

sides  of  the  bar  are  in  contact,  without  showing  signs  of 
fracture.  Wrought  iron  that  breaks  under  this  test  is 
lacking  in  both  strength  and  ductility. 

The  process  of  manufacture,  as  well  as  the  quality  of 
the  pig  iron,  influences  the  strength  of  wrought  iron.  The 
higher  the  percentage  of  carbon  the  greater  is  the  strength. 
Best  iron  is  10  per  cent  stronger  than  ordinary  merchant 
iron  owing  to  the  influence  of  the  second  rolling.  Cold 
rolling  causes  a  marked  increase  in  elastic  limit  and 
ultimate  strength,  but  a  decrease  in  ductility  or  ultimate 
elongation.  Annealing  lowers  the  ultimate  strength,  but 
increases  the  elongation.  Iron  wire,  owing  to  the  process 
of  drawing,  has  a  high  tensile  strength,  sometimes  greater 
than  100  000  pounds  per  square  inch. 

Good  wrought  iron  when  broken  by  tension  shows  a 
fibrous  structure.  If,  however,  it  is  subject  to  shocks 
or  to  repeated  stresses  which  exceed  the  elastic  limit, 
the  molecular  structure  becomes  changed  so  that  the 
fracture  is  more  or  less  crystalline.  The  effect  of  a  stress 
slightly  exceeding  the  elastic  limit  is  to  cause  a  small 
permanent  set,  but  the  elastic  limit  will  be  found  to  be 
higher  than  before.  This  is  decidedly  injurious  to  the 
quality  of  the  material  on  account  of  the  accompanying 
change  in  structure,  and  hence  it  is  a  fundamental  prin- 
ciple that  the  working  unit-stresses  should  not  exceed 
the  elastic  limit.  For  proper  security  indeed  the  allowable 
unit-stress  should  seldom  be  greater  than  one-half  the 
elastic  limit. 

In  a  rough  general  way  the  quality  of  wrought  iron 
may  be  estimated  by  the  product  of  its  tensile  strength 
and  ultimate  elongation,  this  product  being  an  approx- 


ART.  16  STEEL  31 

imate  measure  of  the  work  required  to  produce  rupture. 
Thus  high  tensile  strength  is  not  usually  a  good  quality 
when  accompanied  by  a  low  elongation. 

Prob.  15  A.  What  should  be  the  length  of  a  wrought-iron  bar, 
so  that,  when  hung  at  its  upper  end,  it  will  rupture  there  under  the 
stress  produced  by  its  own  weight? 

Prob.  15  B.  What  is  the  section-area  of  a  bar  of  wrought  iron 
which  weighs  10  pounds  per  linear  foot. 

ART.  16.     STEEL 

Steel  was  originally  produced  directly  from  pure  iron 
ore  by  the  action  of  a  hot  fire,  which  did  not  remove 
the  carbon  to  a  sufficient  extent  to  form  wrought  iron. 
The  modern  processes,  however,  involve  the  fusion  of  the 
ore,  and  the  definition  of  the  United  States  law  is  that 

steel  is  iron  produced  by  fusion  by  any  process,  and 
which  is  malleable."  Chemically,  steel  is  a  compound 
of  iron  and  carbon  generally  intermediate  in  composition 
between  cast  and  wrought  iron,  but  having  a  higher 
specific  gravity  than  either.  The  following  comparison 
points  out  the  distinctive  differences  between  the  three 
kinds  of  iron: 

Per  cent  of  Carbon  Spec.  Grav.             Properties 

Cast  Iron 5       to  2  7.2        Fusible,  not  malleable. 

Steel 1.50  to  0.10  7.8        Fusible  and  malleable. 

Wrought  Iron..  0.30  to  0.05  7.7        Malleable,  not  fusible. 

It  should  be  observed  that  the  percentage  of  carbon 
alone  is  not  sufficient  to  distinguish  steel  from  wrought 
iron;  also,  that  the  mean  values  of  specific  gravity  stated 
are  in  each  case  subject  to  considerable  variation. 

The  three  principal  methods  of  manufacture  are  the 


32  GENERAL   PROPERTIES  CH.  2 

crucible  process,  the  open-hearth  process,  and  the  Bessemer 
process.  In  the  crucible  process  impure  wrought  iron 
or  blister  steel,  with  carbon  and  a  flux,  are  fused  in  a  sealed 
vessel  to  which  air  cannot  obtain  access;  the  best  tool- 
steels  are  thus  made.  In  the  open-hearth  process  pig  iron 
is  melted  in  a  Siemens  furnace,  wrought-iron  scrap  being 
added  until  the  proper  degree  of  carbonization  is  secured. 
In  the  Bessemer  process  pig  iron  is  completely  decar- 
bonized in  a  converter  by  an  air  blast  and  then  recarbon- 
ized  to  the  proper  degree  by  the  addition  of  spiegeleisen. 
The  metal  from  the  open-hearth  furnace  or  from  the 
Bessemer  converter  is  cast  into  ingots,  which  are  rolled  in 
mills  to  the  required  forms.  The  open-hearth  process 
produces  steel  for  guns,  armor  plates,  machinery,  shafts, 
and  for  structural  purposes ;  the  Bessemer  process 
mainly  produces  steel  for  railroad  rails. 

The  physical  properties  of  steel  depend  both  upon  the 
method  of  manufacture  and  upon  the  chemical  composi- 
tion, the  carbon  having  the  controlling  influence  upon 
strength.  Manganese  promotes  malleability  and  silicon 
increases  the  hardness,  while  phosphorus  and  sulphur  tend 
to  cause  brittleness.  The  higher  the  percentage  of  carbon 
within  reasonable  limits  the  greater  is  the  ultimate 
strength  and  the  less  the  elongation. 

A  classification  of  steel  according  to  the  percentage 
of  carbon  and  its  physical  properties  of  tempering  and 
welding  is  as  follows: 

Extra  hard,  1.00  to  0.60%  C.,  takes  high  temper,  but  not  weldable. 
Hard,  0.70  to  0.40%  C.,  temperable,  welded  with  difficulty. 

Medium,      0.50  to  0.20%  C.,  poor  temper,  but  weldable. 
Mild,  0.40  to  0.05%  C.,  not  temperable,  but  easily  welded. 

It  is  seen  that  these  classes  overlap  so  that  there  are  no 


ART.  16  STEEL  33 

distinct  lines  of  demarcation.  The  extra-hard  steels  are 
used  for  tools,  the  hard  steels  for  piston-rods  and  other 
parts  of  machines,  the  medium  steels  for  rails,  tires,  and 
beams,  and  the  mild  or  soft  steels  for  rivets,  plates,  and 
other  purposes. 

The  structural  steel  used  in  bridges  and  buildings 
has  an  ultimate  tensile  strength  of  from  60  000  to  70  000 
pounds  per  square  inch,  with  an  elastic  limit  from  30  000 
to  40  000  pounds  per  square  inch.  The  hard  and  extra- 
hard  steels  are  much  higher  in  strength.  By  the  use  of 
nickel  as  an  alloy  steel  has  been  made  with  an  ultimate 
tensile  strength  of  277  000  and  an  elastic  limit  of  over 
100  000  pounds  per  square  inch. 

The  compressive  strength  of  steel  is  always  higher 
than  the  tensile  strength.  The  maximum  value  recorded 
for  hardened  steel  is  392  000  pounds  per  square  inch. 
The  expense  of  commercial  tests  of  compression  is,  how- 
ever, so  great  that  they  are  seldom  made.  The  shearing 
strength  is  about  three-fourths  of  the  tensile  strength. 

Steel  castings  are  extensively  used  for  axle-boxes, 
cross-heads,  and  joints  in  structural  work.  They  contain 
from  0.25  to  0.50  per  cent  of  carbon,  ranging  in  tensile 
strength  from  60  000  to  100  000  pounds  per  square  inch. 

Steel  has  entirely  supplanted  wrought  iron  for  railroad 
rails,  and  largely  so  for  structural  purposes.  Its  price 
being  the  same,  its  strength  greater,  its  structure  more 
homogeneous,  the  low  and  medium  varieties  are  coming 
more  and  more  into  use  as  a  satisfactory  and  reliable 
material  for  large  classes  of  engineering  constructions. 

Prob.  16  A.     If  steel  costs  3  cents  per  pound  and  nickel  costs 
3 


34  GENERAL    PROPERTIES  CH.  2 

35  cents  per  pound,  what  is  the  minimum  cost  of  a  pound  of  nickel 
steel  which  contains  3.25  per  cent  of  nickel? 

Prob.  16  B.  A  short  steel  piston-rod  is  to  be  designed  to  be  used 
with  a  piston  which  is  20  inches  in  diameter  and  subject  to  a  steam- 
pressure  of  150  pounds  per  square  inch.  If  the  ultimate  strength 
of  the  steel  is  90  000  pounds  per  square  inch,  what  should  be  its 
diameter,  allowing  a  factor  of  safety  of  15? 

ART.  17.     OTHER  MATERIALS 

Common  mortar  is  composed  of  one  part  of  lime  to 
five  parts  of  sand  by  measure.  When  six  months  old  its 
tensile  strength  is  from  15  to  30,  and  its  compressive 
strength  from  150  to  300  pounds  per  square  inch.  Its 
strength  slowly  increases  with  age,  and  it  may  be  con- 
siderably increased  by  using  a  smaller  proportion  of  sand. 

Hydraulic  mortar  is  composed  of  hydraulic  cement 
and  sand  in  varying  proportions.  The  less  the  proportion 
of  sand  the  greater  is  its  strength.  A  common  pro- 
portion is  3  parts  sand  to  1  of  cement,  the  strength  of  this 
being  about  one-fourth  of  the  neat  cement.  The  natural 
cements  are  of  lighter  color,  lower  weight,  and  lesser 
strength  than  the  Portland  cement,  but  they  are  quicker 
in  setting  and  cheaper  in  price.  When  one  week  old, 
neat  natural  cement  has  a  tensile  strength  of  about  125 
and  Portland  cement  about  300  pounds  per  square  inch; 
when  one  year  old  the  tensile  strengths  are  about  300 
and  500  pounds  per  square  inch  respectively.  The  com- 
pressive strength  is  from  8  to  10  times  the  tensile  strength, 
and  it  increases  more  rapidly  with  age. 

Concrete,  composed  of  hydraulic  mortar  and  broken 
stone,  is  an  ancient  material,  having  been  extensively 
used  by  the  Romans.  It  is  mainly  employed  for  founda- 


ART.  17  OTHER   MATERIALS  35 

tions  and  monolithic  structures,  but  in  some  cases  large 
blocks  have  been  made  which  are  laid  together  like 
masonry.  Like  mortar,  its  strength  increases  with  age. 
When  six  months  old  its  mean  compressive  strength 
ranges  from  1000  to  3000  pounds-  per  square  inch,  and 
when  one  year  old  it  is  probably  about  fifty  per  cent 
greater. 

Ropes  are  made  of  hemp,  of  manila,  and  of  iron  or 
steel  wire  with  a  hemp  center.  A  hemp  rope  one  inch 
in  diameter  has  an  ultimate  strength  of  about  6000 
pounds,  and  its  safe  working  strength  is  about  800  pounds. 
A  manila  rope  is  slightly  stronger.  Iron  and  steel  ropes 
one  inch  in  diameter  have  ultimate  strengths  of  about 
36  000  and  50  000  pounds  respectively,  the  safe  working 
strengths  being  6000  and  8000  pounds.  As  a  fair  rough 
rule,  the  strength  of  ropes  may  be  said  to  increase  as  the 
squares  of  their  diameters. 

Aluminum  is  a  silver-gray  metal  which  is  malleable 
and  ductile  and  not  liable  to  corrode.  Its  specific  gravity 
is  about  2.65,  so  that  it  weighs  only  168  pounds  per  cubic 
foot.  Its  ultimate  tensile  strength  is  about  25  000  pounds 
per  square  inch,  and  its  ultimate  elongation  is  also  low. 
Alloys  of  aluminum  and  copper  have  been  made  with  a 
tensile  strength  and  elongation  exceeding  those  of  wrought 
iron,  but  have  not  come  into  use  as  structural  materials. 

Prob.  17  A  Ascertain  the  weight  of  lead  and  brass  per  cubic  foot, 
and  their  ultimate  tensile  strengths. 


Prob.  17  B.  A  bar  of  aluminum  copper,  1%X1%  inches  in  sec- 
tion, breaks  under  a  tension  of  42  800  pounds.  What  tension  will 
probably  bieak  a  bar  of  the  same  material  which  is  1^X2^  inches 
in  section? 


36  GENERAL    PROPERTIES  CH.  2 

ART.   18.     REVIEW  PROBLEMS 

Prob.  ISA  Compute  the  weight  of  a  cast-iron  water-pipe 
12  feet  long,  and  20  inches  inside  diameter,  the  thickness  of  the 
metal  being  1  inch. 

Prob.  18  B.  A  wrought-iron  bar,  1  inch  in  diameter  and  30  feet 
long,  is  hung  at  its  upper  end.  What  load  applied  at  the  lower  end 
will  stress  it  to  the  elastic  limit? 

Prob.  18  C.  Find  the  weight  of  a  wooden  girder  10X12  inches  in 
cross-section  and  16  feet  long. 

Prob.  18  D.  A  cast-iron  column  is  20  feet  long.  Inside  diameter 
of  column  is  8  and  outside  diameter  is  10  inches.  Find  the  weight 
of  this  column. 

Prob.  18  E.  During  a  tension  test  on  a  steel  bar  0.8  inches  in 
diameter  the  following  data  were  obtained:  Maximum  load  on 
specimen  30  000  pounds,  load  at  elastic  limit  23  000  pounds, 
original  length  of  part  tested,  8  inches,  final  length  of  part  tested 
1034  inches,  diameter  at  break  0.32  inches.  Find  the  ultimate 
strength,  elastic  limit,  percent  elongation,  and  percent  reduction 
of  area. 

Prob.  18  F.  A  circular  brick  stack  is  30  feet  high,  its  outside  diam- 
eter being  8  feet  and  inside  diameter  5  feet.  Find  total  weight  of 
stack.  What  is  the  unit-stress  at  the  base  of  the  stack? 

Prob.  18  G.  A  cylinder  head  is  held  in  place  by  six  /'s-inch 
studs.  The  diameter  of  the  cylinder  is  8  inches  and  the  maximum 
steam  pressure  is  200  pounds  per  square  inch.  Find  the  unit  tensile 
stress  at  root  of  threads  if  the  diameter  of  stud  at  root  of  thread 
is  0.507  inches. 

Prob.  18  H.  A  table  with  four  legs  carries  a  load  of  1000  pounds 
uniformly  spread  over  it.  When  the  table  weighs  85  pounds  and 
each  leg  is  1%  inch  in  diameter,  what  is  the  unit-stress  in  the  legs? 

Prob.  18 /.  The  pull  upon  a  IJ^-inch  bolt  passing  through  a 
yellow  pine  girder  is  30  000  pounds.  What  size  of  washer  must 
be  used  so  that  the  unit-stress  on  the  pine  shall  not  exceed  1000 
pounds  per  square  inch? 


ART.  19  THE    PRINCIPLE    OF   MOMENTS  37 


CHAPTER  3 

MOMENTS  FOR  BEAMS 
ART.  19.     THE  PRINCIPLE  OF  MOMENTS 

The  moment  of  a  force  with  respect  to  a  point  is  a 
quantity  which  measures  the  tendency  of  the  force  to 
cause  rotation  about  that  point.  The  moment  is  the 
product  of  the  force  by  the  length  of  its  lever-arm,  the 
lever-arm  being  a  line  drawn  from  the  point  perpen- 
dicular to  the  direction  of  the  force.  Thus  if  P  in  Fig.  7 
is  any  force  and  p  the  length  of  a  perpendicular  drawn  to  it 


from  any  point,  the  product  Pp  is  the  moment  of  the 
force  with  respect  to  that  point.  As  P  is  in  pounds 
and  p  is  in  feet  or  inches,  the  moment  is  a  compound 
quantity  which  is  called  pound-feet  or  pound-inches. 

The  most  important  principle  in  mechanics  is  the 
principle  of  moments.  This  asserts  that  if  any  number 
of  forces  in  the  same  plane  be  in  equilibrium,  the  algebraic 
sum  of  their  moments  about  any  point  in  that  plane  is 
equal  to  zero.  This  principle  results  from  the  meaning 


38  MOMENTS   FOR   BEAMS  CH.  3 

of  the  word  equilibrium,  which  implies  that  the  body 
on  which  the  forces  act  is  at  rest;  and  since  it  is  at  rest 
the  forces  taken  collectively  have  no  tendency  to  turn  it 
around  any  point.  All  experience  teaches  that  the  prin- 
ciple of  moments  is,  indeed,  a  law  of  nature  whose  truth 
is  universal. 

The  point  from  which  the  lever-arms  are  measured ' 
is  often  called  the  'center  of  moments.'  Forces  which 
tend  to  turn  around  this  center  in  the  direction  of  motion 
of  the  hands  of  a  watch  have  positive  moments,  and  those 
which  tend  to  turn  in  the  opposite  direction  have  negative 
moments.  Thus,  in  Fig.  7,  the  numerical  values  of  Pp 
and  Pipi  are  negative,  while  that  of  P^PZ  is  positive.  If 
the  forces  are  in  equilibrium,  the  sum  Pp+Ptfi  has  the 
same  numerical  value  as  Pipi,  or  the  algebraic  sum  of  the 
three  moments  is  zero;  this  will  be  the  case  wherever  the 
center  of  moments  is  taken. 

In  all  investigations  regarding  the  strength  of  beams, 
the  principle  of  moments  is  of  constant  application.  A 
beam  is  a  body  held  in  equilibrium  by  the  downward 
loads  and  the  upward  pressures  of  the  supports.  As  the 
beam  is  at  rest  these  forces  are  in  equilibrium,  and  the 
algebraic  sum  of  their  moments  is  zero  about  any  point 
in  the  plane.  Moreover,  by  further  use  of  the  principle 
of  moments  the  stresses  in  all  parts  of  the  beam  due  to 
the  given  loads  may  be  determined. 

Prob.  19  A.  A  lever  is  5  feet  long  and  the  fulcrum  is  placed  3 
inches  from  one  end.  What  force  will  be  required  at  the  longer  end 
to  lift  1000  pounds  at  the  shorter  end? 

Prob.  19  B.  In  the  above  figure,  the  three  forces  are  in  equilib- 
rium, Pi  being  500  pounds,  P%  being  866  pounds,  and  the  lever-arms 
being  p  =  1.5  feet,  pi=3.5  feet,  p2  =  5.1  feet.  Show  that  the  force 
P  is  4111  pounds. 


ART.  20  REACTIONS  OF  SUPPORTS  39 

ART.  20.     REACTIONS  OF  SUPPORTS 

Let  a  simple  beam  resting  on  two  supports  near  its 
ends  be  subject  to  a  load  P  situated  at  6  feet  from  the 
left  support,  and  let  the  span  be  24  feet  (Fig.  8).  Taking 
the  center  of  moments  at  the  right  support,  the  lever-arm 
of  Ri  is  24  feet,  that  of  P  is  18  feet,  and  that  of  R-2  is 


pi  £f| 


Fig.  8 

0;  then  by  the  application  of  the  principle  of  moments 
fl!X24-PX18  =  0,  or  RV  =  %P.  Again,  taking  the  center 
of  moments  at  the  left  support  the  lever-arm  of  Ri  is  0, 
that  of  P  is  6  feet,  and  that  of  R2  is  24  feet;  then  likewise 
from  the  principle  of  moments  —  -R2X24+PX6  =  0,  or 
Rz  =  ViP.  The  sum  of  these  two  reactions  is  equal  to  P, 
as  should  of  course  be  the  case. 

The  reactions  caused  by  the  weight  of  the  beam  itself 
may  be  found  in  a  similar  manner,  the  uniform  load 
being  supposed  concentrated  at  its  center  of  gravity  in 
stating  the  equations  of  moments.  Thus,  if  the  weight  of 
the  beam  be  W,  the  two  equations  of  moment  are  found 
to  be  B1X24-TTX.12  =  0,  and  -#2X24+TFX12  =  0, 
from  which  R,  =  ^W  and  RZ  =  V2W. 

The  reactions  due  to  both  uniform  and  concentrated 
loads  on  a  simple  beam  may  also  be  computed  in  one 
operation.  As  an  example,  let  there  be  a  simple  beam 
12  feet  long  between  the  supports  and  weighing  35  pounds 
per  linear  foot,  its  total  weight  being  420  pounds  (Fig.  9). 


40  MOMENTS   FOR   BEAMS  CH.  3 

Let  there  be  three  loads  of  300,  60,  and  150  pounds,  placed 
3,  5,  and  8  feet  respectively  from  the  left  support.  To 
find  the  left  reaction  Ri,  the  center  of  moments  is  taken 

,  a»|         601  1501 

H  ---  S--*fr-g-fr  —  S    -*$  ---  4  --  "T 


Fig.  9 


at  the  right  support  and  the  weight  of  the  beam  regarded 
as  concentrated  at  its  middle;  then  the  equation  of 
moments  is 

fl1X!2-420X6-300X9-GOX7-150X4  =  0  • 
from  which  Ri  =  520  pounds.    In  like  manner,  to  find  R2 
the  center  of  moments  is  taken  at  the  left  support;  then 

-#2X12  +  420X6+300X3+GOX5+150X8  =  0 
from  which  .R2  =  410  pounds.     As  a  check  the  sum  of 
Ri  and  R%  is  found  to  be  930  pounds,  which  equals  the 
weight  of  the  beam  and  the  three  loads. 

By  means  of  the  principle  of  moments  other  prob- 
lems relating  to  reactions  of  beams  may  also  be  solved. 
For  instance,  if  a  simple  beam  12  feet  long  weighs  30 
pounds  per  linear  foot  and  carries  a  load  of  600  pounds, 
where  should  this  load  be  put  so  that  the  left  reaction 
may  be  twice  as  great  as  the  right  reaction?  Here  let 
x  be  the  distance  from  the  left  support  to  the  load;  let 
RI  be  the  left  reaction  and  R2  the  right  reaction.  Then 
taking  the  centers  of  moments  at  the  right  and  left  sup- 
port in  succession  there  are  found 

R1=  180+50  (12  -x),    fl2=180+50z 
and  placing  Ri  equal  to  2R2  there  results  x  =  2.8  feet. 


ART.  21  BENDING   MOMENTS  41 

Prob.  20  A.  A  beam  weighing  30  pounds  per  linear  foot  rests 
upon  two  supports  16  feet  apart.  A  weight  of  400  pounds  is  placed 
at  5  feet  from  the  left  end,  and  one  of  600  pounds  is  placed  at  8  feet 
from  the  right  end.  Find  the  reactions  due  to  the  total  load. 

Prob.  20  B.  A  wooden  girder,  8X10  inches  in  section-area  and 
18  feet  between  supports,  carries  a  uniformly  distributed  load  of 
500  pounds  per  linear  foot  for  a  distance  of  8  feet  from  the  left  end. 
The  remaining  10  feet  carry  a  uniformly  distributed  load  of  700 
pounds  per  linear  foot.  Find  the  reactions  at  the  supports. 

ART.  21.     BENDING  MOMENTS 

The  'bending  moment'  at  any  section  of  a  beam  is  the 
algebraic  sum  of  the  moments  of  all  the  vertical  forces 
on  the  left  of  that  section.  It  is  a  measure  of  the  tendency 
of  those  forces  to  cause  rotation  around  that  point.  At 
the  ends  of  a  simple  beam  there  are  no  bending  moments, 
but  at  all  other  sections  they  exist,  and  the  greater  the 
bending  moment  the  greater  are  the  horizontal  stresses 
in  the  beam,  these  stresses  in  fact  being  produced  by  the 
bending  moment. 

For  example,  let  a  beam  30  feet  long  have  three  loads 
of  100  pounds  each,  situated  at  distances  of  8,  12,  and  22 
feet  from  the  left  support  (Fig.  10).  By  the  method  of 


Fig.  10 

the  previous  article  the  left  reaction  Ri  is  160  pounds 
and  the  right  reaction  Rz  is  140  pounds.  For  a  section 
4  feet  from  the  left  support  the  bending  moment  is  160X4 
=  640  pound-feet,  and  for  a  section  at  8  feet  from  the 


42  MOMENTS    FOR   BEAMS  CH.  3 

left  support  the  bending  moment  is  160X8=  1280  pound- 
feet.  For  a  section  10  feet  from  the  left  support  there 
are  two  vertical  forces  on  the  left  of  the  section,  160 
acting  up  and  100  acting  down,  so  that  the  bending 
moment  is  160X10-100X2-1400  pound-feet.  For  a 
section  at  the  middle  of  the  beam  the  bending  moment  is 
160  X 15  - 100  X  7  - 100  X  3  =  1400  pound-feet.  For  a  sec- 
tion under  the  third  load  the  bending  moment  is,  in  like 
manner,  1120  pound-feet,  and  for  a  section  at  3  feet  from 
the  right  support  it  is  420  pound-feet.  The  vertical 
ordinates  underneath  the  beam  represent  the  values  of 
these  bending  moments,  and  the  diagram  thus  formed 
shows  how  the  bending  moments  vary  throughout  the 
length  of  the  beam. 

For  a  simple  beam  of  span  I  and  uniformly  loaded  with 
w  pounds  per  linear  unit,  each  reaction  is  Vzwl.  For  any 
section  distant  x  from  the  left  support  (Fig.  11),  the 
bending  moment  is  */<>wlXx— wxXVix,  where  the  lever- 


Fig.  11 

arm  of  the  reaction  is  x  and  the  lever-arm  of  the  load 
wx  is  Vsx.  If  w  is  80  pounds  per  linear  foot  and  I  is  30 
feet,  the  bending  moment  at  any  section  is  then  1200z— 
40z2.  For  3=10  feet,  the  bending  moment  is  8000 
pound-feet;  for  a;  =15  feet  it  is  9000  pound-feet;  for 
x  —  20  feet  it  is  8000  pound-feet,  and  so  on.  The  diagram 
shows  the  distributions  of  moments  throughout  the 


ART.  22  RESISTING  MOMENTS  43 

beam,  and  it  can  be  demonstrated  that  the  curve  joining 
the  ends  of  the  ordinates  is  the  common  parabola. 

When  a  beam  is  loaded  both  uniformly  and  with  con- 
centrated loads,  the  bending  moments  for  all  sections 
may  be  found  in  a  similar  manner.  The  maximum 
bending  moment  indicates  the  point  where  the  beam  is 
under  the  greatest  horizontal  stresses;  this  will  usually 
be  found  near  the  middle  of  the  beam  and  often  under 
one  of  the  concentrated  loads.  For  simple  beams  resting 
on  two  supports  at  their  ends  all  the  bending  moments  are 
positive.  It  may  further  be  noted  that  if  the  vertical 
forces  on  the  right  of  the  section  be  used,  the  same  numer- 
ical values  will  be  found  for  the  bending  moments. 

Prob.  21  A.  Two  locomotive  wheels,  six  feet  apart  and  each 
weighing  20  000  pounds,  roll  over  a  beam  of  27  feet  span.  Find  the 
greatest  reaction  which  can  be  caused  by  these  wheels. 

Prob.  21  B.  A  simple  beam  of  12  feet  span  weighs  60  pounds 
per  linear  foot,  and  has  a  load  of  150  pounds  at  8  feet  from  the 
left  end.  Compute  the  bending  moments  for  sections  distant  2, 
4,  6,  8,  10  feet  from  the  left  support,  and  construct  the  diagram  of 
bending  moments. 

ART.  22.     RESISTING  MOMENTS 

Suppose  a  simple  beam  to  be  cut  by  an  imaginary, 
vertical  plane  MN  and  the  portion  on  the  right  of  that 
plane  to  be  removed  (Fig.  12) .  In  order  that  the  remaining 
part  may  be  in  equilibrium,  forces  must  be  applied  to 
the  section;  in  the  figure  horizontal  forces  are  shown, 
and  these  represent  the  horizontal  stresses  in  the  section. 
The  reaction  and  loads  on  the  left  of  M N  together  with 
the  stresses  acting  on  that  section  constitute  a  system  of 
forces  in  equilibrium.  The  algebraic  sum  of  the  moments 


44 


MOMENTS   FOR   BEAMS 


CH.  3 


of  the  reaction  and  loads  with  respect  to  the  point  D 
is  the  bending  moment  for  the  section  MN,  the  value  of 
which  may  be  found  by  the  methods  of  the  last  article. 
This  bending  moment  tends  to  turn  the  beam  in  a  clock- 


?" 

D 

nr~ 

i 

c 

_r 

E^ 

Fig.  12 

wise  direction  about  D.  and  it  is  balanced  by  the  sum  of  the 
moments  of  the  stresses  acting  on  MN,  which  turn  it  in 
the  opposite  direction.  Hence 

Bending  Moment  =  Resisting  Moment. 

It  will  now  be  shown  how  an  expression  for  the  second 
term  of  this  equation  can  be  obtained. 

It  is  found  by  experiment  that  there  is  a  certain  line 
CD  on  the  side  of  the  beam  which  does  not  change  in 
length  under  the  bending,  and  hence  there  is  no  horizontal 
stress  upon  it.  Below  this  neutral  line  the  fibers  in  a 
simple  beam  are  found  to  be  elongated,  and  above  it  they 
are  shortened;  thus  the  stresses  below  the  neutral  line 
are  tension  and  those  above  it  are  compression.  A  neutral 
line  like  CD  also  obtains  in  all  longitudinal  vertical  sec- 
tions of  the  beam.  There  is,  in  fact,  a  'neutral  surface' 
extending  throughout  the  entire  width  of  the  beam,  and 
the  intersection  of  this  neutral  surface  with  any  section 
area  gives  a  line  CC,  which  is  called  the  'neutral  axis'  of 
that  section. 


ART.  22  RESISTING  MOMENTS  45 

It  is  also  found  by  experiment  that  the  horizontal 
stresses  in  any  section  increase  uniformly  from  the  neutral 
axis  to  the  top  and  bottom  of  the  beam,  provided  the 
elastic  limit  of  the  material  is  not  exceeded.  Thus,  if  S 
is  the  horizontal  unit-stress  at  the  upper  or  lower  side 
of  the  beam  in  Fig.  11,  the  unit-stress  halfway  between 
that  side  and  the  neutral  axis  is  %S.  Also,  let  c  be  the 
distance  from  the  neutral  axis  to  the  upper  or  lower  side 
of  the  beam,  and  z  be  any  distance  less  than  c,  then  the 
horizontal  unit-stress  at  the  distance  z  is  S(z/c). 

'Resisting  Moment'  is  the  term  used  to  denote  the 
algebraic  sum  of  the  moments  of  all  the  horizontal 
stresses  in  a  section  with  respect  to  its  neutral  axis.  Let  a 
(Fig.  12)  be  any  small  elementary  area  of  the  section  at  a 
distance  z  from  the  neutral  axis.  The  unit-stress  on  this 
small  area  is  S(z/c)  and  hence  the  total  stress  on  it  is 
aS(z/c).  The  moment  of  this  stress  with  respect  to  the 
point  D  is  aS(z/c)  multiplied  by  its  lever-arm  z,  or  aS 
(z/c)z.  Hence 

Moment  of  stress  on  a  =  (S/c)az2 

and  the  resisting  moment  is  the  algebraic  sum  of  all  these 
elementary  moments  for  all  possible  values  of  z,  or  since 

S/c  is  a  constant, 

Sf  Sf 

Resisting  Moment =-(alzl2+a2z22+a3  z32+  — )  =-2az2 

C  C 

Here   the  notation    2az2  is  used  to  denote  the  quantity 

aiZi2+02Z22+ The  letter  S  is  not  a  factor,  but  a  symbol 

which  indicates  the  process  of  summation   and  it  should 
be  called  'summation  of  all  values  of.' 

This  quantity  Zaz2  is  called  the  '  moment  of  inertia '  of 
the  cross-section  of  the  beam,  and  it  will  be  shown  in 


46  MOMENTS   FOR  BEAMS  CH.   3 

Art.  24  how  its  value  is  found.  The  moment  of  inertia 
is  designated  by  /,  and  hence 

SI 
Resisting  Moment  =  — 

This  expression  for  the  resisting  moment  is  applicable  to 
all  kinds  of  cross-sections.  It  is  shown  above  that  the 
bending  moment  for  any  section  is  equal  to  the  resisting 
moment  for  that  section;  letting  M  be  the  value  of  the 
bending  moment,  found  as  in  Art.  21,  then 


is  a  general  formula  applicable  to  all  kinds  of  beams. 
This  formula  will  be  constantly  used  in  the  next  chapter. 

The  term  '  moment  of  inertia  '  has  no  reference  to  inertia 
when  applied  to  plane  surfaces,  as  is  here  the  case.  It  is 
merely  a  name  for  the  quantity  2az2,  and  this  quantity 
is  found  by  multiplying  each  elementary  area  by  the 
square  of  its  distance  from  the  given  axis  and  taking  the 
sum  of  the  products.  As  z2  is  always  positive  whether  z 
be  positive  or  negative,  Saz2  or  the  moment  of  inertia  /  is 
always  positive.  If  all  the  elementary  areas  be  taken  as 
equal  and  n  be  their  number,  then  na  =  A,  the  total  area 
of  the  section.  Hence  /  =  Saz2  =  a  2z2  =  A  2z2/n.  Desig- 
nate the  average  of  all  the  values  of  z2  by  r2,  then  2z2/n  =  r- 
and  thus  I  =  A  r2  is  another  definition  of  moment  of  inertia. 
This  is  a  constant  which  depends  only  on  the  size  of  the 
section  area  and  its  arrangement  with  respect  to  the 
neutral  surface. 

Prob.  22.  In  the  above  ngure  let  MD  and  DN  be  each  6  inches 
and  let  the  width  of  the  beam  CC  be  8  inches.  If  the  tensile  unit- 
stress  S  on  the  bottom  of  the  beam  is  600  pounds  per  square  inch, 
the  compressive  unit-stress  on  the  top  of  the  beam  is  also  600  pounds 


ART.  23  CENTERS  OF  GRAVITY  47 

per  square  inch.    Show  that  the  total  tensile  stress  is  14  400  pounds, 
and  that  the  total  compressive  stress  is  also  14  400  pounds. 

ART.  23.     CENTERS  OF  GRAVITY 

The  center  of  gravity  of  a  plane  surface  is  that  point 
upon  which  a  thin  sheet  of  cardboard,  having  the  same 
shape  as  the  given  surface,  can  be  balanced  when  held 
in  a  horizontal  position.  In  the  investigation  of  beams 
its  section  area  is  the  given  surface,  and  it  is  required 
to  know  the  distances  from  the  top  or  bottom  of  the 
section  to  the  center  of  gravity.  The  letter  c  will  be 
used  to  denote  these  distances  when  they  are  equal, 
and  the  longest  of  these  distances  when  they  are  unequal. 

For  a  square,  rectangle,  or  circle,  whose  depth  is  d, 
it  is  evident  that  c  =  %d.  Also  for  a  section  of  I  shape, 
where  the  upper  and  lower  flanges  are  equal  in  size,  it 
is  plain  that  c  =  y%d. 

For  the  T  section,  shown  on  the  left  of  Fig.  13,  the 
distance  c  is  greater  than  %d,  and  its  value  is  to  be  found 
by  using  the  principle  of  moments.  If  the  width  of  the 
horizontal  flange  is  4  inches  and  its  thickness  1%  inches, 
the  area  of  the  flange  is  5  square  inches;  if  the  height  of 


Fig.  13 

the  vertical  web  is  6  inches  and  its  thickness  1  inch,  the 
area  of  the  web  is  6  square  inches.  The  total  area  of  the 
cross-section  is  then  11  square  inches.  Now  if  this  sec- 
tion is  a  thin  sheet  held  in  a  horizontal  plane,  the  weights 


48  MOMENTS    FOR   BEAMS  CH.  3 

of  the  two  parts  and  the  whole  are  represented  by  5,  6, 
and  11.  With  respect  to  an  axis  at  the  end  of  the  web 
the  lever-arms  of  these  weights  are  6%  inches,  3  inches, 
and  c  inches;  the  equation  of  moments  then  is 

5X6%+6X3-llXc  =  0 

from  which  the  value  of  c  is  found  to  be  4.65  inches.  For 
the  channel  section,  shown  on  the  right  of  Fig.  13,  the 
same  method  is  to  be  followed  as  for  the  T. 

The  method  of  moments  may  thus  be  applied  to  areas 
as  well  as  to  forces.  If  a  be  any  area  and  z  the  distance 
of  its  center  of  gravity  from  an  axis,  the  product  az  is 
called  the  static  moment  of  the  area.  The  algebraic 
sum  of  the  static  moments  of  all  parts  of  the  figure  is 
represented  by  2az,  the  summation  of  the  values  a\Zi, 
o^Zz,  a&z,  etc.  If  A  is  the  total  section  area,  then 


is  a  general  expression  of  the  method  of  finding  the  dis- 
tance c.  If  the  axis  is  taken  within  the  section,  some  of 
the  z's  are  negative,  and  if  the  axis  passes  through  the 
center  of  gravity  of  the  section,  then  the  quantity  Saz 
is  zero. 

When  the  cross-section  is  bounded  by  curved  lines, 
as  in  a  railroad  rail,  it  is  to  be  divided  up  into  small 
rectangles  and  the  value  of  a  be  found  for  each;  the  sum 
of  all  the  a's  is  A,  and  then  by  the  above  method  the 
value  of  c  is  computed.  For  the  various  rolled  shapes 
found  in  the  market  the  values  of  c  are  thus  determined 
by  the  manufacturers  and  published  for  the  information 
of  engineers. 


ART.  23  CENTERS  OF  GRAVITY  49 

Triangular  beams  are  never  used,  but  it  is  often  con- 
venient to  remember  that  for  any  triangle  whose  depth 
is  d  the  value  of  c  is  %d. 

For  the  angle  section,  shown  in  Fig.  14,  the  center  of 
gravity  usually  lies  without  the  section,  and  there  are 
two  values  of  c,  called  Ci  and  Cz,  to  be  determined.  Let 
the  thickness  of  each  leg  be  %  inches,  the  length  of  the 
long  leg  be  6  and  that  of  the 
short  leg  be  4  inches.  The 
area  of  the  long  leg,  including 
the  lower  corner,  is  6X%  = 
4.5  square  inches,  and  its 
center  of  gravity  is  3  inches 
below  the  axis  AA  and  3  % 
inches  to  the  right  of  axis  BB. 
The  area  of  the  short  leg,  excluding  the  corner,  is 
=  2.4375  square  inches,  and  its  center  of  gravity  is  5% 
inches  below  the  axis  AA  and  1%  inches  to  the  right  of 
the  axis  BB.  Then,  as  the  total  area  of  the  section  is 
6.9375  square  inches,  the  equation  of  moments  with 
respect  to  the  axis  A  A  is 

6.9375  d  =  4.5X3+2.4375  X  5.625, 

and  then  ci=  3.92  inches.     Also  the  equation  of  moments 
with  respect  to  the  axis  BB  is 

6.9375  C2  =  4.5X3.625+2.4375X  1.625, 
from  which  C2  =  2.92  inches. 

Prob.  23  A.  For  Fig.  13  let  c=6  and  ci  =3  inches.  If  the  unit 
stress  S  at  the  top  of  the  web  is  6400  pounds  per  square  inch,  what 
is  the  unit-stress  Si  on  the  lower  side  of  the  flange? 

Prob.  23  B.     A  deck-beam  used  in  buildings  has  a  rectangular 
4 


50 


MOMENTS    FOR   BEAMS 


CH.  3 


flange  4X%  inches,  a  rectangular  web  5XK  inches,  and  an  ellip- 
tical head  which  is  1  inch  in  depth  and  whose  area  is  1.6  square 
inches.  Find  the  distance  of  the  center  of  gravity  from  the  top  of 
the  head. 

ART.  24.     MOMENTS  OF  INERTIA 

The  moment  of  inertia  of  a  plane  surface  with  respect 
to  an  axis  is  the  sum  of  the  products  obtained  by  mul- 
tiplying each  elementary  area  by  the  square  of  its  distance 
from  that  axis.  In  the  discussion  of  beams  the  axis  is 
always  taken  as  passing  through  the  center  of  gravity 
of  the  cross-section  and  parallel  to  the  top  and  bottom 
lines  of  the  cross-section.  Let  /  be  this  moment  of  inertia 
as  in  Art.  22,  its  value  is  to  be  found  by  determining 
the  quantity  2a22,  the  summation  of  all  the  values  a\z\, 
a2z22,  a3z32,  etc. 

To  find  7  for  a  rectangle  of  breadth  b  and  depth  d, 
let  CC  be  the  axis  through  the  center  of  gravity  and 
parallel  to  b  (Fig.  15).  Let  the  elementary  area  a  be  a 


Fig.  15 

small  strip  EE  parallel  to  CC  and  at  a  distance  z  from  it. 
Let  a  line  gh  be  drawn  parallel  to  the  depth  d  of  the 
rectangle,  and  normal  to  gh  let  lines  be  drawn  equal  to 
the  squares  of  z;  thus  ee  is  the  square  of  CE,  and  gg  is 
the  square  of  CG.  Now  the  elementary  product  az~  is  the 


ART.  24 


MOMENTS    OF   INERTIA 


51 


elementary  area  EE  multiplied  by  the  ordinate  ee;  hence 
Zaz2  is  represented  by  a  solid  standing  on  bd  whose  variable 
height  is  shown  by  the  shaded  area  ghhcg.  But  the 
volume  of  this  solid  is  the  product  of  its  length  b  and  this 
shaded  area.  The  curve  ceg  is  a  parabola  because  each 
line  ee  is  the  square  of  the  corresponding  altitude  ce; 
accordingly  the  shaded  area  is  one-third  of  ghhg.  But 
gh  is  equal  to  d,  and  gg  is  equal  to  (%d)2;  thus  the  shaded 
area  is  represented  by  Va-d.^id?,  or  Mod3.  Hence 


is  the  moment  of  inertia  of  a  rectangle  about  an  axis 
through  its  center  of  gravity  and  parallel  to  its  base. 

The  moment  of  inertia  is  a  compound  quantity  resulting 
by  multiplying  an  area  by  the  square  of  a  distance;  it 
thus  contains  the  linear  unit  four  times.  If  6  =  3  inches 
and  d  =  4  inches,  then  7=16  inches4,  or  the  numerical 
unit  of  7  is  biquadratic  inches. 

Moments  of  inertia  when  referred  to  the  same  axis 
can  be  added  or  subtracted  like  any  other  qualities 


t> 

JL    Fig-16 

which  are  of  the  same  kind.  Thus,  let  there  be  a  hollow 
rectangular  section  whose  outside  depth  and  breadth 
are  6  and  d  and  whose  inside  depth  and  breadth  are 
61  and  di,  the  thickness  of  the  metal  being  the  same 
throughout  (Fig.  16).  Then  the  moment  of  inertia  of 


52  MOMENTS    FOR   BEAMS  CH.  3 

this  section  is  found  by  subtracting  the  moment  of  inertia 
of  the  inner  rectangle  from  that  of  the  outer  one,  or 


is  the  moment  of  inertia  for  the  rectangular  section  whose 
area  is  bd  —  bidi. 

For  the  common  I  beams  whose  flanges  are  equal 
the  same  method  applies.  Let  b  be  the  width  of  the 
flanges  and  d  the  total  depth  of  the  section  shown  on  the 
right  of  Fig.  16;  also  let  t  be  the  thickness  of  the  web 
and  tr  the  thickness  of  the  flanges.  The  moment  of 
inertia  of  the  area  (6  —  f)(d—  2ti)  is  then  to  be  subtracted 
from  the  moment  of  inertia  of  the  area  bd,  or 


is  the  moment  of  inertia  for  the  I  section. 


Fig.  17 

For  the  "J"  section  the  distance  c  from  the  end  of  the 
web  to  the  axis  through  the  center  of  gravity  must  first 
be  computed  by  the  method  of  the  last  article.  Then  ci, 
the  distance  from  the  outside  of  the  flange  to  the  axis, 
is  also  known  (Fig.  17).  Let  6  be  the  breadth  of  the 
flange  and  ti  its  thickness,  and  let  t  be  the  thickness  of 
the  web.  Then  the  moment  of  inertia  of  the  area  tc  is 
one-half  of  that  of  a  rectangle  of  depth  2c,  or  ^X  Via'(2c)3, 
which  is  i/^c3;  also  the  moment  of  inertia  of  the  area 
bci  is  one-half  of  that  of  a  rectangle  of  depth  2ci.  Adding 


ART.  24  MOMENTS  OF  INERTIA  53 

these  together  and  subtracting  the  moment  of  inertia 
of  the  area  (b  —  0(ci  —  ti),  there  results 


which  is  the  moment  of  inertia  for  the  "f  section.  The 
same  formula  applies  to  the  LJ  section  if  t  is  the  thick- 
ness of  the  two  webs. 

The  above  formulas  for  I  and  "f  sections  are  correct 
for  cast-iron  beams  Vhere  the  corners  are  but  little 
rounded.  For  wrought-iron  and  steel  beams,  however, 
the  flanges  are  not  usually  of  uniform  thickness,  and 
all  the  corners  are  rounded  off  by  curves,  so  that  the 
formulas  are  not  strictly  correct;  for  such  shapes  the 
numerical  values  of  the  moments  of  inertia  for  all  the 
sections  in  the  market  are  published  by  the  manufac- 
turers, so  that  it  is  not  necessary  for  engineers  to  compute 
them.  (See  Art.  33.) 

The  moment  of  inertia  of  a  circle  with  respect  to  its 
diameter  as  an  axis  is  /  =  %47rd4  where  d  is  the  length 
of  the  diameter. 

The  moment  of  inertia  of  a  plane  surface  with  respect 
to  any  axis  is  equal  to  the  moment  of  inertia  with  respect 
to  a  parallel  axis  through  its  center  of  gravity  plus  the 
area  of  the  surface  multiplied  by  the  square  of  the  dis- 
tance between  the  two  axes.  Thus,  let  A  be  the  area  of  a 
surface  (Fig.  18),  /  its  moment  of  inertia  with  respect 
to  an  axis  CC  through  the  center  of  gravity,  and  h  the 
distance  to  another  parallel  axis  DD.then  the  moment  of 
inertia  of  the  surface  with  Yespect  to  DD  is  Ii  =  I-\-Ah2. 
This  principle  is  used  to  find  the  moment  of  inertia  for 
compound  and  built-up  sections,  as  illustrated  in  the 
following  example. 


54 


MOMENTS   FOB    BEAMS 


CH.  3 


For  the  section  of  a  bridge-post,  shown  in  Fig.  19,  the 
area  of  each  12-inch  channel  and  its  moment  of  inertia 
with  respect  to  an  axis  through  its  center  of  gravity 
are  given  as  A  =7.35  square  inches  and  /=  144.0  inches4. 
For  the  plate,  which  is  16  X  %  inches,  the  values  are 
A  =  6.00  square  inches  and  I  =  0.007  inches4.  From  Art.  21 


D- 


-D 


Fig.  18 


Fig  19 


the  distance  c  to  the  center  of  gravity  of  the  compound 
section  is  found  to  be  7.99  inches.  Hence  for  each  channel 
/i=1.99  inches,  and  for  the  plate  /i  =  4.57  inches.  The 
moment  of  inertia  for  the  entire  built-up  section  is  then 

71  =  2(144.0+7.35X1.992)+0.007+6.00X4.572  =  471.4. 

In  this  chapter  the  fundamental  applications  of  mo- 
ments to  be  used  in  discussing  beams  have  been  presented, 
and  it  is  now  possible  to  take  up  the  subject  and  give  the 
theory  of  equilibrium  of  beams  clearly  and  logically,  so 
that  the  student  may  undertake  practical  problems  in 
the  most  satisfactory  manner. 

Prob.  24.  A  steel  I  beam  weighing  80  pounds  per  linear  foot  is 
24  inches  deep,  its  flanges  being  7  inches  wide  and  J/%  inches  mean 
thickness,  while  the  web  is  0.5  inches  thick.  The  moment  of  inertia 
stated  by  the  manufacturer  is  2088  inches4.  Compute  it  by  the 
formula  here  given. 

ART.   25.  REVIEW  PROBLEMS 

Prob.  25  A,  Three  men  carry  a  stick  of  timber,  two  taking  hold 
at  a  common  point  and  one  at  one  of  the  ends.  Where  should  be 


ART.  25  REVIEW  PROBLEMS  55 

the  common  point  so  that  each  man  may  carry  one-third  of  the 
weight? 

Prob.  25  B.  Compute  the  bending  moments  under  each  con- 
centrated load  for  Fig.  9,  taking  the  weight  of  the  beam  into  account. 

Prob.  25  C.  The  two  bases  of  a  trapezoid  are  8  and  5  inches, 
and  its  height  is  4  inches.  Find  the  center  of  gravity. 

Prob.  25  D.  For  a  solid  circular  section  the  moment  of  inertia 
with  respect  to  an  axis  through  the  center  is  %t7rd4.  Find  the 
moment  of  inertia  for  a  hollow  circular  section  with  outside  diameter 
di  and  inside  diameter  ck. 

Prob.  25  E.  A  simple  beam  of  16  feet  span  weighs  60  pounds 
per  linear  foot  and  has  a  concentrated  load  of  500  pounds  at  a  dis- 
tance of  4  feet  from  the  left  end.  Compute  the  bending  moments 
for  several  sections  throughout  the  beam  and  construct  the  diagram 
of  moments. 

Prob.  25  F.  Locate  both  gravity  axes  of  a  standard  steel  channel 
8  inches  deep,  the  average  thickness  of  the  web  being  0.22  inches, 
average  thickness  of  flange  0.40  inches,  and  width  of  flanges  2.26 
inches. 

Prob.  25  G.  Compute  the  moments  of  inertia  of  the  channel 
section  with  respect  to  each  of  the  axes  found  in  the  last  problem. 

Prob.  25  H.  Find  the  moment  of  inertia  of  a  circle  3  inches  in 
diameter.  Also  the  moment  of  inertia  of  that  circle  with  respect  to 
another  axis  in  the  same  plane,  the  shortest  distance  from  the  center 
of  the  circle  to  that  axis  being  5  inches. 

Prob.  25  J.  A  timber  cantilever  4X6  inches  in  section  projects 
6  feet  out  of  a  wall.  What  load  must  be  put  upon  it  so  that  the 
greatest  shearing  stress  shall  be  120  pounds  per  square  inch? 

Prob.  25  K.  Show  that  the  moment  of  inertia  of  a  rectangle 
with  respect  to  an  axis  passing  through  its  base  is 


56  CANTILEVER  AND  SIMPLE  BEAMS  Cfl.  4 

CHAPTER  4 

CANTILEVER   BEAMS   AND  SIMPLE   BEAMS 
ART.  26.     DEFINITIONS  AND  PRINCIPLES 

A  simple  beam  is  a  bar  resting  upon  supports  at  its 
ends,  and  is  the  kind  most  commonly  in  use.  A  cantilever 
beam  is  a  bar  resting  on  one  support  at  the  middle,  or 
if  a  part  of  a  beam  projects  out  from  a  wall  or  beyond  a 
support  this  part  is  called  a  cantilever  beam.  In  a  simple 
beam  the  lower  part  is  under  tension  and  the  upper  part 
under  compression;  in  a  cantilever  beam  the  reverse 
is  the  case.  Unless  otherwise  stated,  all  beams  will  be 
regarded  as  having  the  section  area  uniform  throughout 
the  entire  length. 

Since  a  beam  is  at  rest  the  internal  stresses  in  any  section 
hold  in  equilibrium  the  external  forces  on  each  side  of 
that  section.  Thus,  if  a  beam  be  imagined  to  be  cut  apart 


i 


Fig.  20 

and  the  two  parts  separated,  as  in  Fig.  20,  forces  must 
necessarily  be  required  to  prevent  the  parts  from  falling. 
These  internal  forces  or  stresses  may  be  resolved  into 
horizontal  and  vertical  components.  The  horizontal 
components  are  stresses  of  tension  and  compression,  while 


ART.  26 


DEFINITIONS   AND    PRINCIPLES 


57 


the  vertical  components  add  together  and  form  a  stress  V 
known  as  the  resisting  shear. 

Each  side  of  the  beam  is  held  in  equilibrium  by  the 
vertical  forces  and  stresses  that  act  upon  it.  The  vertical 
forces  are  the  reaction  and  the  loads,  the  stresses  are  the 
horizontal  ones  of  tension  and  compression,  and  the 
vertical  one  of  shear.  The  sum  of  all  the  horizontal 
tensile  stresses  must  be  equal  to  the  sum  of  all  the  hori- 
zontal compressive  stresses,  or  otherwise  there  would  be 
longitudinal  motion.  The  sum  V  of  the  vertical  stresses 
must  equal  the  algebraic  sum  of  the  reaction  and  loads, 
or  otherwise  there  would  be  motion  in  an  upward  or 
downward  direction.  Lastly,  the  sum  of  the  moments  of 
the  stresses  in  the  section  must  equal  the  sum  of  the 
moments  of  the  vertical  forces,  or  otherwise  there  would 
be  rotation.  These  statical  principles  apply  to  each  part 
into  which  the  beam  is  supposed  to  be  divided. 

It  is  found  by  experiment  that  the  fibers  on  one  side 
of  the  beam  are  elongated  and  on  the  other  shortened, 
while  between  is  a  neutral  surface,  which  is  unchanged 

-S 


Fig.  21 

in  length.  It  is  also  found  that  the  amount  of  elongation 
or  shortening  of  any  fiber  is  directly  proportional  to  its 
distance  from  the  neutral  surface.  Hence,  if  the  elastic 
limit  is  not  surpassed,  the  stresses  are  also  proportional 
to  their  distances  from  the  neutral  surface  (Fig.  21). 


58  CANTILEVER   AND    SIMPLE   BEAMS  CH.  4 

From  the  above  it  can  be  shown  that  the  neutral  axis 
passes  through  the  center  of  gravity  of  the  cross-section. 
For,  if  S  be  the  unit-stress  on  the  remotest  fiber  and  c  its 
distance  from  the  neutral  axis,  then  the  unit-stress  at  the 
distance  z  is  Sz/c,  and  the  total  stress  on  an  elementary 
area  a  is  aSz/c.  The  algebraic  sum  of  all  the  horizontal 
stresses  in  the  section  then  is  (S/c)  Zaz,  where  Saz  denotes 
the  summation  of  all  the  values  a\z\,  a2z2,  etc.  From  the 
above  statical  principles  this  sum  must  be  zero,  and  it 
hence  follows  that  2az  must  be  zero;  that  is,  the  sum  of 
the  moments  of  the  elementary  areas  is  zero  with  respect 
to  the  neutral  axis.  Hence  the  neutral  axis  passes  through 
the  center  of  gravity,  for  the  center  of  gravity  is  that 
point  upon  which  the  surface  can  be  balanced,  or  it  is  that 
point  for  which  1,az  =  0. 

Prob.  26  A.  An  I  beam  which  is  20  feet  long  weighs  700  pounds 
and  the  area  of  its  cross-section  is  10.29  square  inches.  What  is 
the  kind  of  material? 

Prob.  26  B.  Let  Oi=2,  a2=2.5,  a3  =  2.7  square  inches,  and 
Zj  =  +3.5,  22  =  +1.5,  z?=  —2.6  inches.  Does  the  axis  pass  through 
the  center  of  gravity  of  c^ +02+03? 

ART.  27.     RESISTANCE  TO  SHEARING 

When  a  beam  is  short  it  sometimes  fails  by  shearing 
in  a  vertical  section  near  one  of  the  supports.  The  force 
that  produces  this  shearing  is  the  resultant  of  all  the 
vertical  forces  on  one  side  of  the  section.  Thus,  in  the 
simple  beam  of  the  first  diagram  (Fig.  22)  this  resultant 
is  the  reaction  minus  the  weight  of  the  beam  between 
the  reaction  and  the  section;  in  the  cantilever  beam  of 
the  second  diagram  it  is  the  loads  and  the  weight  of  the 
beam  on  the  left  of  the  section. 


ART.  27  RESISTANCE  TO  SHEARING  59 

'Vertical  shear'  is  the  name  given  to  the  algebraic 
sum  of  all  the  vertical  forces  on  the  left  of  the  section 
under  consideration.  Thus  in  the  first  diagram  of  Fig.  22, 
if  the  reaction  is  6000  pounds,  the  vertical  shear  V  just  at 


the  right  of  the  support  is  6000  pounds.  If  the  beam 
weighs  100  pounds  per  linear  foot,  the  vertical  shear  at 
a  section  one  foot  from  the  support  and  on  the  left  of 
the  single  load  is  5900  pounds.  Again  in  the  second 
diagram  of  Fig.  22,  if  the  beam  weighs  100  pounds 
per  linear  foot  and  if  each  concentrated  load  is  800 
pounds,  and  the  distance  from  the  end  to  the  section 
shown  is  4  feet,  the  vertical  shear  in  that  section  is  2000 
pounds. 

It  is  seen  from  these  illustrations  that  in  a  simple 
beam  the  greatest  vertical  shear  is  at  the  supports,  and 
that  in  a  cantilever  beam  it  is  at  the  wall.  Only  these 
sections,  then,  need  be  investigated  in  a  solid  beam.  For 
a  simple  beam  of  length  I  and  carrying  w  pounds  per 
linear  unit,  the  greatest  vertical  shear  is  the  reaction 
Y^wl.  For  a  cantilever  beam  of  length  I,  the  greatest 
vertical  shear  due  to  uniform  load  is  the  total  weight  wl. 


60  CANTILEVER   AND    SIMPLE    BEAMS  CH.  4 

The  vertical  shear  V  produces  in  the  cross-section 
an  equal  shearing  stress.  If  A  is  the  section  area  and  S 
the  shearing  unit-stress  acting  over  that  area,  then 

V  =  AS,  S  =  ~  A  =  ^  (3) 

are  the  equations  similar  to  (1)  of  Art.  1;  these  are  used 
for  the  practical  computations  regarding  shear  in  solid 
beams. 

For  example,  consider  a  steel  I  beam  weighing  250 
pounds  per  yard  and  12  feet  long,  over  which  roll  three 
locomotive  wheels  4  feet  apart  and  each  bearing  14  000 
pounds  (Fig.  23).  The  greatest  shear  will  occur  when 


Fig.  23 

one  wheel  is  almost  at  the  support  as  shown  in  the  figure. 
By  Art.  20  the  reaction  is  found  to  be  28  500  pounds, 
and  this  is  the  greatest  vertical  shear  V.  By  Art.  9  the 
area  of  the  cross-section  is  found  to  be  24.5  square  inches. 
Then  the  shearing  unit-stress  in  the  section  is 

28  500 
S  =  — — -  =  1160  pounds  per  square  inch 

^4.O 

which  is  a  low  working  unit-stress  for  steel. 

As  a  second  example,  consider  a  wooden  cantilever 
beam  which  projects  out  from  a  bridge  floor  and  supports 
a  sidewalk.  Let  it  be  6  inches  wide,  8  inches  deep,  and 
7  feet  long,  and  let  the  maximum  load  that  comes  upon  it 
be  7500  pounds.  The  vertical  shear  at  the  section  where 
it  begins  to  project  is  then  7590  pounds,  or  the  load  that 


ART.  28  RESISTANCE  TO  BENDING  61 

it  carries  plus  its  own  weight.  As  the  section  area  is  48 
square  inches,  the  shearing  unit-stress  is  a  little  less  than 
160  pounds  per  square  inch.  The  factor  of  safety  against 
shearing  is  hence  about  19  (Art.  6),  so  that  the  security 
is  ample. 

It  is  indeed  only  in  rare  instances  that  solid  beams  of 
uniform  cross-section  are  subject  to  dangerous  stresses 
from  shearing.  Beams  almost  universally  fail  by  tearing 
apart  under  the  horizontal  tensile  stresses,  and  hence 
the  following  articles  will  be  devoted  entirely  to  the  con- 
sideration of  these  bending  stresses. 

Prob.  27  A.  A  simple  beam  of  cast  iron  is  3 X3  inches  in  section 
and  5^  feet  long  between  supports.  Besides  its  own  weight,  it  is 
to  carry  a  load  of  4000  pounds  at  the  middle  and  a  load  of  1000 
pounds  at  23/2  feet  from  the  left  end.  Find  the  factor  of  safety 
against  shearing. 

Prob.  27  B.  On  a  simple  beam  12  feet  long  there  are  two  loads, 
each  600  pounds,  one  at  3  feet  from  the  left  end,  and  one  at  3  feet 
from  the  right  end.  Find  the  vertical  shear  due  to  these  loads  for 
a  section  near  one  of  the  supports,  and  also  for  any  section  between 
the  loads. 

ART.  28.     RESISTANCE  TO  BENDING 

In  Art.  27  it  was  shown  that  the  resisting  moment 
of  the  internal  stresses  in  any  section  is  equal  to  the 
bending  moment  of  the  external  forces  on  each  side  of 
the  section.  Art.  21  explains  how  to  find  the  bending 
moment,  which  hereafter  will  be  designated  by  the  letter 
M.  In  Art.  22  an  expression  for  the  resisting  moment 
is  derived.  Therefore 

f  =  A/  (4) 

is  the  fundamental  formula  for  the  discussion  of  the 


62  CANTILEVER   AND   SIMPLE   BEAMS  CH.  4 "} 

flexure  of  beams,  provided  the  elastic  limit  is  not  exceeded. 
Here  S  is  the  unit-stress  of  tension  or  compression  on  the 
top  or  bottom  of  the  beam,  c  is  the  vertical  distance  of  S 
from  the  center  of  gravity  of  the  cross-section,  and  7 
is  the  moment  of  inertia  of  the  cross-section.  Art.  23 
explains  how  to  find  c,  and  Art.  24  shows  how  /  is  deter- 
mined, S  is  often  called  the  'fiber  unit-stress,'  meaning 
thereby  the  greatest  horizontal  unit-stress. 

This  formula  shows  that  S  varies  directly  with  M, 
that  is,  the  greatest  tensile  or  compressive  stress  in  the 
beam  occurs  at  the  section  where  M  has  its  maximum 
value.  For  a  simple  beam  under  uniform  load  the  bending 
moment  M  at  any  section  distant  x  from  the  left  support  is, 
as  shown  in  Art.  21, 

M = y2wi.  x  -  wx .  y2x = y#n  (ix  -  x2) 

and  if  x  =  3^Z,  this  gives  M  =  ^wl2  as  the  maximum 
bending  moment;  or  if  W  be  the  total  load  wl,  this  may  be 
written  as  M  =  l/gWl.  When  concentrated  loads  are  on  a 
simple  beam  the  maximum  bending  moment  must  usually 
be  found  by  trial;  it  will  generally  be  under  one  of  those 
loads. 

For  a  cantilever  beam  of  length  I  the  maximum  bending 
moment  always  occurs  at  the  wall  (Fig.  24).  For  a 
uniform  load  of  w  per  linear  unit  the  bending  moment 
at  a  section  distant  x  from  the  end  is  the  load  wx  into 
its  lever-arm  %x,  and  this  is  negative  as  it  tends  to  pro- 
duce rotation  in  a  direction  opposite  to  that  of  the  hands 
of  a  watch  (Art.  19).  Thus,  for  any  section  M  =  —  %wx2, 
and  when  x  becomes  equal  to  I  the  maximum  value  is 
M  —  —  }/2wl2.  The  negative  sign  shows  merely  the  direc- 
tion in  which  rotation  tends  to  occur,  and  when  using 


ART.  28 


RESISTANCE   TO   BENDING 


formula  (4)  the  value  of  M  is  to  be  inserted  without  sign. 
The  diagram  of  bending  moments  for  this  case  is  a  para- 
bola, since  M  increases  as  the  square  of  x. 


Fig.  24 

For  concentrated  loads  on  a  cantilever  beam  the 
bending  moment  M  is  —  FIX  until  x  passes  beyond  the 
second  load;  then  M=—P1x—Pz(x—p)  where  p  is  the 
distance  between  the  two  loads.  Thus  the  diagram  of 
bending  moments  is  composed  of  straight  lines  (Fig.  25), 


Fig.  25 

the  maximum  M  occurs  when  x  becomes  equal  to  Z,  and 
its  value  is  -Pil-P2(l-p). 

It  may  be  noted  that  the  only  difference  in  stating 
moment  equations  for  a  cantilever  beam  and  for  a  simple 
beam  lies  in  the  fact  that  for  the  former  there  is  no 


64  CANTILEVER   AND    SIMPLE   BEAMS  CH.  4 

reaction  at  the  left  end.  A  cantilever  beam  is  hence  really 
simpler  than  a  simple  beam,  as  no  reactions  need  be 
computed. 

Prob.  28-4.  A  cantilever  beam  has  a  load  of  800  pounds  at  its 
end,  and  is  also  uniformly  loaded  with  125  pounds  per  linear  foot; 
its  length  is  5  feet.  Compute  the  bending  moments  for  five  sections, 
one  foot  apart,  and  construct  the  diagram  of  bending  moments. 

Prob.  28  B.  A  simple  beam  weighing  60  pounds  per  linear  foot 
is  13  feet  in  span  and  has  a  load  of  1000  pounds  at  the  middle. 
Compute  the  maximum  bending  moment. 

ART.  29.     SAFE  LOADS  FOR  BEAMS 

A  safe  load  for  a  beam  is  one  that  produces  a  tensile 
or  compressive  unit-stress  which  is  safe  according  to 
the  principles  set  forth  in  Chapter  1.  To  find  such  a  safe 
load  for  a  given  beam  the  safe  value  of  S  is  to  be  assumed 
from  those  principles.  Then  in  formula  (4)  the  values  of 
7  and  c  are  known.  The  maximum  bending  moment  M 
is  to  be  expressed  in  terms  of  the  unknown  load,  and  thus 
an  equation  is  derived  from  which  the  load  is  found. 

For  example,  let  a  wooden  cantilever  beam  be  2  inches 
wide,  3  inches  deep,  and  72  inches  long,  and  let  it  be 
required  to  find  what  load  P  at  the  end  will  produce  a 
unit-stress  S  of  800  pounds  per  square  inch.  Here  the 
maximum  value  of  M  is  PX72.  From  Art.  23  the  value 
of  c  is  \Yi  inches  and  from  Art.  24  the  value  of  /  is  4^ 
inches4.  Then  by  (4)  of  Art.  28, 


c  1.5 

from  which  P  is  found  to  be  33%  pounds. 
Again,  let  a  simple  beam  of  cast  iron  be  3  inches  wide, 


ART.  29  SAFE  LOADS  FOR  BEAMS  65 

4  inches  deep,  and  36  inches  long,  and  let  it  be  required 
to  find  what  uniform  load  will  produce  a  unit-stress  of 
2000  pounds  per  square  inch.  Here  let  w  be  the  uniform 
load  per  linear  inch;  the  total  load  is  wl,  each  reaction  is 
%wl,  and  the  maximum  bending  moment  M  is  }/%wlz. 
The  value  of  c  is  2  inches,  and  that  of  7  is  16  inches4. 
Then  since  I  is  36  inches, 

SI  ^  2000  X 16 
c  ~         2 

from  which  w  =  98.8  pounds  per  linear  inch,  and  hence 
the  total  uniform  safe  load  that  can  be  put  on  the  beam 
is  about  3560  pounds. 

The  student  should  notice  that  in  using  formula  (4) 
all  lengths  must  be  expressed  in  the  same  unit.  If  the 
length  of  a  beam  is  given  in  feet  it  must  be  reduced  to 
inches  for  use  in  the  formula,  because  S,  I,  and  c  are 
expressed  in  terms  of  inches.  Formula  (4)  cannot  be 
used  to  find  the  load  that  will  rupture  a  beam,  except 
in  the  manner  indicated  in  Art.  66. 

Prob.  29  A .  A  steel  I  beam  7  inches  deep  and  weighing  22  pounds 
per  foot  has  for  the  moment  of  inertia  of  its  cross-section  52.5 
inches4,  and  it  is  to  be  used  as  a  simple  beam  with  a  span  of  18  feet. 
What  load  P  can  it  carry  when  the  greatest  unit-stress  S  is  required 
to  be  12  000  pounds  per  square  inch? 

Prob.  29  B.  WThat  safe  uniformly  distributed  load  can  be  carried 
by  an  oak  beam,  4  inches  wide  and  6  inches  deep,  having  a  span 
of  16  feet,  if  the  greatest  unit-stress  is  not  to  exceed  1000  pounds 
per  square  inch? 

Prob.  29  C.     A  wooden  beam,   10X12  inches  in  section  area, 
projects  6  feet  from  the  wall  of  a  building.    What  safe  load  can  be 
suspended  from  the  end  of  the  beam  so  that  there  shall  be  a  factor 
of  safety  of  10? 
5 


66  CANTILEVER  AND    SIMPLE    BEAMS  CH.  4 

ART.  30.    INVESTIGATION  OF  BEAMS 

To  investigate  a  beam  acted  upon  by  given  loads  the 
greatest  unit-stress  S  produced  by  those  loads  is  to  be 
found  from  formula  (4).  From  the  given  dimensions  of 
the  beam  7  an$  c  are  known,  from  the  given  loads  the 
maximum  value  of  M  is  to  be  found;  then 
«  Me 

s=T 

is  the  equation  for  computing  the  value  of  S.  Then 
by  the  rules  of  Chapter  1  the  degree  of  security  of  the 
beam  is  to  be  inferred.  As  formula  (4)  is  deduced  under 
the  laws  of  elasticity,  it  fails  to  give  reliable  values  of  S 
when  the  elastic  limit  is  exceeded. 

For  example,  consider  a  cast-iron  u  section  which  is 
used  as  a  simple  beam  with  a  span  of  6  feet,  and  upon 
which  there  is  a  total  uniform  load  of  SO  000  pounds. 
Let  the  total  depth  be  16  inches,  the  total  width  12 
inches,  the  thickness  of  the  flange  2  inches,  and  the 
thickness  of  the  webs  1  inch.  By  Art.  28  the  greatest 
value  of  M  is  at  the  middle  of  the  beam,  this  being 
Ys  X  80  000  X  6  X  12  =  720  000  pound-inches.  By  Art.  23 
the  value  of  c  is  found  to  be  10.7  inches.  By  Art.  24  the 
value  of  I  is  found  to  be  1292  inches4.  Then, 

„    720000X10.7 

"  =  -  10no  -  =  5960  pounds  per  square  inch. 


This  is  the  compressive  unit-stress  in  the  end  of  the  web 
when  the  beam  is  placed  in  the  LJ  position,  as  is  usually 
the  case  in  buildings.  On  the  base  of  the  beam  the  tensile 
unit-stress  is  about  half  this  value,  since  ci  is  about  one- 
half  of  c.  Thus  under  the  compressive  stress  the  beam 


ART.  31  DESIGN  OF  BEAMS  67 

has  a  factor  of  safety  of  about  15,  and  under  the  tensile 
stress  it  has  a  factor  of  safety  of  about  7.  As  the  least 
factor  of  safety  for  cast  iron  should  be  10,  the  beam 
has  not  the  full  degree  of  security  required  by  the  best 
practice. 

Prob.  30-4.  A  piece  of  wooden  scantling  2  inches  square  and  18 
feet  long  is  hung  horizontally  by  a  rope  at  each  end  and  a  student 
weighing  175  pounds  stands  upon  it.  Is  it  safe? 

Prob.  30  B.  A  floor  is  supported  by  3  X  8-inch  wooden  joists  of 
16  feet  span  spaced  18  inches  apart.  When  this  floor  carries  a  total 
load  of  200  pounds  per  square  foot,  what  is  the  factor  of  safety  of 
the  joists? 

ART.  31.     DESIGN  OF  BEAMS 

The  design  of  a  beam  consists  in  determining  its  size 
when  the  loads  and  its  length  are  given.  The  allowable 
working  unit-stress  S  is  first  assumed  according  to  the 
requirements  of  practice.  From  the  given  loads  the 
maximum  bending  moment  M  is  then  computed.  Thus 
in  formula  (4)  everything  is  known  except  7  and  c,  and 

L-M 

c     S 

is  an  equation  which  must  be  satisfied  by  the  dimensions 
to  be  selected. 

For  a  rectangular  beam  of  breadth  b  and  depth  d  the 
value  of  c  is  %d,  and  the  value  of  I  is  %2  b<&.  Thus  the 
equation  above  becomes 

,  _    6M 


and  if  either  b  or  d  be  assumed  the  other  can  be  computed. 
For  example,  let  it  be  required  to  design  a  rectangular 
wooden  beam  for  a  total  uniform  load  of  80  pounds,  the 


68  CANTILEVER  AND    SIMPLE   BEAMS  CH.  4 

beam  to  be  used  as  a  cantilever  with  a  length  of  6  feet, 
and  the  working  value  of  S  to  be  800  pounds  per  square 
inch.  Here  the  maximum  value  of  M  is  80X3  =  240 
pound-feet  =  2880  pound-inches.  Thus  bd2  =  2  1  .6  inches3. 
If  6  is  taken  as  1  inch,  d  =  4.65  inches;  if  6  is  2  inches, 
d=3.29  inches;  if  6  is  3  inches,  d  =  2.Q8  inches.  With  due 
regard  to  sizes  readily  found  in  the  market  3X3  inches 
are  perhaps  good  proportions  to  adopt. 

Prob.  31-4.  A  simple  cast-iron  beam  of  14  feet  span  carries  a  load 
of  10  000  pounds  at  the  middle.  If  its  width  is  4  inches,  find  its 
depth  for  a  factor  of  safety  of  10;  also  find  its  width  for  a  depth  of 
12  inches. 

Prob.  31  B.  A  yellow  pine  beam  of  20  feet  span  is  to  carry  a  uni- 
formly distributed  load  of  500  pounds  per  linear  foot  with  a  factor 
of  safety  of  9.  The  depth  of  the  beam  is  to  be  1  ^  times  the  breadth. 
Find  the  dimensions  of  the  beam. 

ART.  32.     COMPARATIVE  STRENGTHS 

The  strength  of  a  beam  is  measured  by  the  load  it 
can  carry  with  a  given  unit-stress  S.  Let  it  be  required  to 
investigate  the  relative  strengths  of  the  four  following 
cases  : 

1st.  A  cantilever  loaded  at  the  end  with  W. 

2d.    A  cantilever  loaded  uniformly  with  W. 

3d.    A  simple  beam  loaded  at  the  middle  with  W. 

4th.  A  simple  beam  loaded  uniformly  with  W. 

Let  I  be  the  length  in  each  case,  and  the  cross-section  be  of 
breadth  b  and  depth  d.  Then  c  =  ^d,  and  I  =  Vi2bd?. 
Then,  from  Art.  28,  and  formula  (4), 


For  1st,    M=  Wl      and  W  =  -~ 


ART.  32  COMPARATIVE  STRENGTHS  69 

For  2d,     M  =  Y2Wl  and  W  =  ^f 
For  3d,     M 
For  4th,   M= 


Hence  the  comparative  strengths  of  the  four  cases  are 
as  the  numbers  1,  2,  4,  8;  that  is,  if  four  such  beams  are 
of  equal  size  and  length  and  of  the  same  material,  the 
second  is  twice  as  strong  as  the  first,  the  third  is  four 
times  as  strong,  and  the  last  is  eight  times  as  strong  as 
the  first. 

From  these  equations  the  following  important  laws 
regarding  rectangular  beams  are  derived  : 

The  strength  varies  directly  as  the  breadth  and  directly  as  the 

square  of  the  depth. 

The  strength  varies  inversely  as  the  length. 
A  beam  is  twice  as  strong  under  a  distributed  load  as  under  an 

equal  concentrated  load. 

The  second  and  third  of  these  laws  apply  also  to  beams 
having  cross-sections  of  any  shape. 

The  reason  why  rectangular  beams  are  placed  with 
the  longest  dimension  vertical  is  now  seen  to  be  that 
the  strength  increases  in  a  faster  ratio  with  the  depth 
than  with  the  breadth.  If  the  breadth  is  doubled  the 
strength  is  doubled;  if  the  depth  is  doubled  the  strength 
is  four  times  as  great  as  before. 

A  beam  is  said  to  be  'fixed'  at  its  end  when  the  end  is 
fastened  in  a  wall  in  such  a  manner  that  that  end 
remains  horizontal.  The  following  are  the  maximum 
bending  moments  for  such  beams: 


70  CANTILEVER   AND    SIMPLE   BEAMS  CH.  4 

One  end  fixed,  load  W  at  middle,     M  =  ^Wl 
One  end  fixed,  uniform  load  W,        M=  }/%Wl 
Both  ends  fixed,  load  W  in  middle,  M=y%Wl 
Both  ends  fixed,  uniform  load  W,     M  =  yl2Wl 

It  is  thus  seen  that  beams  fixed  at  their  ends  are  stronger 
than  simple  beams  similarly  loaded,  for  under  a  given 
unit-stress  S  they  will  carry  a  greater  load  W.  Moment 
diagrams  for  fixed  beams  are  given  in  Art.  57 

An  'overhanging  beam'  is  shown  in  Fig.  26,  the  dis- 
tance between  the  supports  being  I,  one  support  being 


k 


Fig.  26 

at  the  left  end  and  the  other  support  being  at  a  distance 
m  from  the  right  end.  Under  a  uniform  load  of  w  per 
linear  unit,  the  left  reaction  is  Ri  =  ]^w(l—mi/l),  the 
maximum  positive  moment  is  M  =  Y^R^Iw  and  the  max- 
imum negative  moment  is  M  =  ^4wmz.  For  instance,  let 
1=10  and  m  =  Q  feet,  and  w  =  30  pounds  per  linear  foot; 
then  the  total  weight  of  the  beam  is  480  pounds,  the  left 
reaction  is  96  pounds,  the  maximum  positive  bending 
moment  is  153.6,  and  the  maximum  negative  bending 
moment  is  540  pound-feet. 

For  a  load  P  placed  between  the  supports  of  the  over- 
hanging beam  in  Fig.  26,  the  reactions  Ri  and  #2  are 
exactly  the  same  as  for  a  simple  beam  and  the  maximum 


ART.  33 


STEEL  I   BEAMS 


71 


moment  is  l/iPl.  For  a  load  P  placed  at  the  end  of  the 
overhanging  arm,  the  reactions  are  Ri=  —Pm/l  and  R2  = 
+P(1—  m/l),  while  the  greatest  bending  moment  is  Pm. 

Prob.  32  A.  Show  that  a  beam  3  inches  wide,  6  inches  deep,  and 
4  feet  long  is  nine  times  as  strong  as  a  beam  2  inches  wide,  4  inches 
deep,  and  10%  feet  long. 

Prob.  32  B.  Compute,  without  using  the  above  formulas,  the 
reactions  for  an  overhanging  beam  where  the  distance  between  the 
supports  is  9  feet  and  the  overhanging  arm  is  3  feet,  the  beam 
weighing  40  pounds  per  linear  foot. 

ART.  33.     STEEL  I  BEAMS 

Wrought-iron  rolled  beams  have  been  much  used  in 
bridge  and  building  construction,  but  now  medium-steel 
beams  are  almost  exclusively  employed.  The  ultimate 
tensile  strength  of  such  steel  will  be  taken  as  65  000 
pounds  per  square  inch,  and  its  elastic  limit  as  35  000 
pounds  per  square  inch,  in  the  solution  of  examples  and 


problems  hereafter  given.  These  beams  are  manufactured 
in  about  thirteen  different  depths,  and  of  each  depth 
there  are  several  different  sizes  or  weights,  so  that  designers 


72 


CANTILEVER   AND    SIMPLE    BEAMS 


CH.  4 


have  a  large  variety  from  which  to  select.    In  the  following 
table  only  the  heaviest  and  lightest    sections  of  each 

TABLE  8.     STEEL  I  BEAMS 


Depth 
Inches 

Weight  per 
Foot 

Pounds 

Section 
Area 

A 
Sq.  Inches 

Moment  of 
Inertia 

I 
Inches* 

Section 
Modulus 

/ 

e 
Inchess 

Moment  of 
Inertia 

/' 

Inches4 

24 

100 

29.4 

2380 

198 

48.6 

24 

80 

23.5 

2088 

174 

42.9 

20 

75 

22.1 

1269 

127 

30.2 

20 

65 

19.1 

1170 

117 

27.9 

18 

70 

20.6 

921 

102 

24.6 

18 

55 

15.9 

796 

88.4 

21.2 

15 

55 

15.9 

511 

68.1 

17.1 

15 

42 

12.5 

442 

58.9 

14.6 

12 

35 

10.3 

228 

38.0 

10.1 

12 

31H 

9.3 

216 

36.0 

9.50 

10 

40 

11.8 

159 

31.7 

9.50 

10 

25 

7.4 

122 

24.4 

6.89 

9 

35 

10.3 

112 

24.8 

7.31 

9 

21 

6.3 

85.0 

18.9 

5.16 

8 

25  % 

7.50 

68.4 

17.1 

4.75 

8 

18 

5.33 

56.9 

14.2 

3.78 

7 

20 

5.88 

42.2 

12.1 

3.24 

7 

15 

4.42 

36.2 

10.4 

2.67 

6 

17M 

5.07 

26.2 

8.73 

2.36 

6 

12M 

3.61 

21.8 

7.27 

1.85 

5 

14% 

4.34 

15.2 

6.08 

1.70 

5 

9% 

2.87 

12.1 

4.84 

1.23 

4 

i(% 

3.09 

7.1 

3.55 

1.01 

4 

7y2 

2  21 

6.0 

3.00 

0.77 

3 

7y2 

2.21 

2.9 

1.93 

0.60 

3 

5y2 

1.63 

2.5 

1.71 

0.46 

depth  are  given.     The  proportions  of  one  of  the  sizes  of 
the  two  6-inch  beams  are  shown  in  Fig.  27,  the  outer  line 


ART.  33  STEEL  I  BEAMS  73 

on  the  right-hand  side  indicating  the  heavier  section  and 
the  other  one  the  lighter  section. 

In  Table  8  the  moments  of  inertia  /  in  the  fourth 
column  are  those  about  an  axis  through  the  centers  of 
gravity  and  perpendicular  to  the  web,  and  are  those  to 
be  used  in  all  beam  computations.  The  values  /'  given 
in  the  last  column  are  with  respect  to  an  axis  through 
the  center  of  gravity  but  parallel  to  the  web;  these  are 
for  use  in  the  next  chapter  in  the  discussion  of  struts. 

The  quantity  I/c  is  often  called  the  'section  modulus' 
as  it  contains  all  the  dimensions  of  the  cross-section. 
The  process  of  selecting  an  I  section  depends  merely 
on  finding  a  value  of  I/c  which  corresponds  to  the  value 
of  M/S,  as  shown  in  Art.  31;  hence  for  convenience  these 
values  are  tabulated  in  the  fifth  column  of  the  table. 

For  example,  an  I  beam  in  a  floor  is  to  have  20  feet 
span  and  to  carry  a  uniform  load  of  13  500  pounds; 
what  size  is  to  be  selected?  The  bending  moment  is 
M  =  y8 X 13  500 X 20 X  12  =  405  000  pound-inches;  and  the 
working  unit-stress  S  should  be  %X65  000  pounds  per 
square  inch.  Then  from  formula  (4), 

7     405  000 


c       13  000 


=  31.2  inches3 


and  hence,  from  the  table,  the  heavy  10-inch  beam  should 
be  used. 

Prob.  33-4.  A  heavy  15 -inch  steel  I  beam  of  simple  span  carries 
a  uniform  load  of  42  net  tons.  Find  its  factor  of  safety  if  the  span 
is  6  feet;  also  if  the  span  is  9  feet. 

Prob.  33  B.  A  steel  I  beam  of  25  feet  span  is  to  carry  a  uniformly 
distributed  load  of  1000  pounds  per  linear  foot.  In  addition  there 


74 


CANTILEVER  AND   SIMPLE    BEAMS 


CH.  4 


is  a  concentrated  load  of  6000  pounds  at  10  feet  from  the  left  end. 
Find  the  proper  size  of  I  beam  to  be  used. 

ART.  34.     BEAMS  OF  UNIFORM  STRENGTH 

The  beams  thus  far  discussed  have  been  of  uniform 
section  throughout  their  entire  length.  As  the  bending 
moments  are  small  near  the  ends  of  the  beam  the  unit- 
stress  S  is  there  also  small,  and  hence  more  material 
is  used  than  is  really  needed.  A  beam  of  uniform  strength 
is  one  so  shaped  that  the  unit-stress  S  is  the  same  at  all 
parts  of  the  length. 

For  a  cantilever  beam  loaded  with  P  at  the  end,  the 
bending  moment  at  any  distance  x  from  the  end  is  Px. 


Fig.  28. 

If  the  section  is  rectangular,  formula  (4)  reduces  to 
V6Sbd2  =  Pxf  in  which  P  and  S  are  constant.  If  6  is  made 
the  same  throughout,  then 

***j* 

and  therefore  d2  must  vary  directly  as  x.  If  x  =  l,  the 
value  of  d  is  the  depth  di  at  the  wall,  and  accordingly 
6P/Sb=*di2/l;  hence  the  equation  becomes 


d  =  . 


ART.  34 


BEAMS  OF  UNIFORM  STRENGTH 


75 


Thus,  if  x  =  %l,  d  =  Vsdi;  if  x  =  %l,  d=%di,  and  so  on. 
As  the  squares  of  the  depths  vary  with  the  distances  from 
the  end,  the  curve  of  the  side  of  the  beam  should  be  the 
common  parabola  (Fig.  28). 

For  a  rectangular  cantilever  beam  uniformly  loaded  with 
w  per  linear  unit  the  bending  moment  M  is  %wx*  and 

I 


Fig.  29 


formula  (4)  becomes 
uniform  throughout,  then 


If  the  breadth  is 


^j 

oo 


Here,  if  x  =  l,  the  value  of  d  is  the  depth  di  at  the  wall, 
and  thus  3w/Sb  =  dlP/P.    Accordingly, 


gives  the  depth  for  any  value  of  x,  and  it  shows  that 
the  elevation  of  the  beam  should  be  a  triangle  (Fig.  29). 

The  vertical  shear  near  the  end  of  the  beam  modifies 
slightly  the  form  near  the  end.  Thus  for  the  first  case 
above,  if  Sf  be  the  working  shearing  unit-stress  there 
must  be  a  section  at  the  end  whose  area  A  is  equal  at 
least  to  P/S'. 

Prob.  34.  A  simple  beam  of  uniform  strength  is  to  be  designed 
to  carry  a  heavy  load  P  at  the  middle.  If  d\  be  the  depth  at  the 
middle,  show  that  the  depths  at  distances  O.I/,  0.21,  Q.31,  and  QAl 
should  be  0.45d,,  0.63di,  0.77di,  and  0.89*. 


76  CANTILEVER   AND    SIMPLE    BEAMS  CH.    4 

ART.  35.     REVIEW  PROBLEMS 

Prob.  35  A.  Locate  the  neutral  axis  for  a  "f"  section  which  is 
3X3  inches  and  %  inches  thick. 

Prob.  35  B.  A  timber  4X6  inches  in  section  projects  6  feet  out 
of  a  wall.  What  load  must  be  put  upon  it  so  that  the  greatest 
shearing  stress  shall  be  120  pounds  per  square  inch? 

Prob.  35  C.  A  simple  wooden  beam,  8  inches  wide,  9  inches  deep, 
and  14  feet  in  span,  carries  two  equal  loads,  one  being  2.5  feet 
at  the  left  and  the  other  2.5  feet  at  the  right  of  the  middle.  Find 
these  loads  so  that  the  factor  of  safety  of  the  beam  shall  be  8. 

Prob.  35  D.  A  simple  wooden  beam,  3  inches  wide,  4  inches  deep, 
and  16  feet  span,  has  a  load  of  150  pounds  at  the  middle.  Compute 
its  factor  of  safety. 

Prob.  35  E.  A  simple  beam  of  structural  steel,  %  inches  deep 
and  16  feet  span,  is  subject  to  a  rolling  load  of  500  pounds.  What 
must  be  its  width  in  order  that  the  factor  of  safety  may  be  6? 

Prob.  35  F.  Compare  the  strength  of  a  joist,  3X8  inches,  when 
laid  with  long  side  vertical  with  that  when  it  is  laid  with  short  side 
vertical. 

Prob.  35  G.  Compare  the  strength  of  a  light  9-inch  steel  I  beam 
with  that  of  a  wooden  beam  8  inches  wide  and  12  inches  deep,  the 
span  being  the  same  for  both. 

Prob.  35  H.  A  cast-iron  cantilever  beam  is  to  be  4  feet  long, 
3  inches  wide,  and  to  carry  a  load  of  15  000  pounds  at  the  end. 
Find  the  proper  depths  for  every  foot  of  length,  using  3000  pounds 
per  square  inch  for  the  horizontal  unit-stress  and  4000  for  the 
vertical  shearing  unit-stress. 

Prob.  35  «/.  The  wooden  girders  of  a  floor  are  10X12  inches  in 
cross-section,  25  feet  span  and  16  feet  apart.  The  floor  carries  a 
load  of  100  pounds  per  square  foot.  Find  the  maximum  unit-stress 
at  the  middle  of  the  girders. 

Prob.  35  K.  A  steel  pin,  8  inches  long  and  3  inches  in  diameter, 
is  arranged  like  a  simple  beam  to  carry  a  load  of  10  000  pounds 
at  the  middle.  Find  the  maximum  flexural  unit-stress. 


ART.  36  GENERAL  PRINCIPLES  77 

CHAPTER  5 
COLUMNS  OR  STRUTS 

ART.  36.     GENERAL  PRINCIPLES 

A  bar  under  compression  whose  length  is  greater  than 
about  ten  times  its  thickness  is  called  a  column  or  a  strut. 
For  shorter  lengths  the  case  is  one  of  direct  compression 
where  the  rules  of  Art.  5  apply.  For  the  short  specimen 
failure  occurs  by  the  shearing  or  splintering  of  the  material. 
For  the  strut  or  column,  however,  failure  generally 
occurs  by  a  sidewise  bending;  this  induces  bending 
stresses,  so  that  the  phenomena  of  stress  are  more  complex 
than  in  a  beam. 

Wooden  and  cast-iron  columns  are  usually  square  or 
round,  and  are  sometimes  built  hollow.  Wrought-iron 
columns  are  made  by  riveting  together  channels,  plates, 
and  angle-irons.  It  is  clear  that  a  square  or  round  section 
is  preferable  to  a  rectangular  one,  since  then  the  tendency 
to  bend  is  the  same  in  all  directions.  For  a  rectangular 
section  the  bending  will  evidently  occur  in  a  plane  parallel 
to  the  shorter  side  of  the  rectangle;  thus  in  investigating 
such  a  column  the  depth  d  is  this  shorter  side  instead  of 
the  longer  one,  as  in  beams.  When  a  single  I  beam  is 
used  as  a  column  it  tends  to  bend  in  a  plane  parallel 
to  the  flanges,  and  hence  the  moment  of  inertia  to  be 
used  in  this  discussion  is  /',  which  is  given  in  the  last 
column  of  the  table  in  Art.  33,  the  axis  for  this  coinciding 
with  the  middle  line  of  the  web. 

If  a  short  prism  whose  section    area  is  A  is  loaded 


78  COLUMNS   OR   STRUTS  CH.  5 

with  the  weight  P,  the  unit-stress  is  P/A,  and  this  is 
uniformly  distributed  over  the  area  A.  For  a  column, 
however,  this  is  not  the  case;  while  the  mean  unit-stress 
is  still  P/A,  the  unit-stress  on  the  concave  side,  if  bending 
occurs,  may  be  very  much  greater  than  P/A.  The 
longer  the  column  the  greater  is  this  unit-stress  on  the 
concave  side  liable  to  become,  and  hence  a  long  column 
cannot  carry  so  large  a  load  as  a  short  one. 

There  are  three  ways  of  arranging  the  ends  of  columns 
(Fig.  30).    Class  (a)  includes  those  with  'round  ends'  or 


Fig.  30 

those  having  their  ends  hinged  on  pins.  Class  (6)  includes 
those  with  one  end  round  and  the  other  fixed;  the  piston- 
rod  of  a  steam-engine  is  of  this  type.  Class  (c)  includes 
those  having  fixed  ends;  these  are  used  in  bridge  and 
building  constructions.  The  figure  here  given  is  a  sym- 
bolical representation,  and  is  not  intended  to  imply 
that  the  ends  of  the  columns  are  necessarily  enlarged  in 
practice.  It  is  found  by  experiment  that  class  (c)  is 
stronger  than  (6),  and  that  (6)  is  stronger  than  (a). 

Prob.  36.  In  a  certain  test  wrought-iron  tubes  2.37  inches  in 
outer  diameter  and  having  a  section  area  of  1.08  square  inches 
were  used.  A  tube  8  feet  long  failed  under  24  800  pounds  and  a 


ART.  37  RADIUS  OF  GYRATION  79 

tube  3*4  feet  long  failed  under  38200  pounds.     What  load  would 
be  required  to  cause  failure  for  a  tube  only  6  or  8  inches  long? 

ART.  37.     RADIUS  OF  GYRATION 

In  the  discussion  of  columns  a  quantity  r,  called  'radius 
of  gyration  of  the  cross-section,'  is  frequently  used. 
It  is  defined  to  be  that  quantity  whose  square  is  equal 
to  the  least  moment  of  inertia  of  the  cross-section  divided 
by  the  area  of  that  cross-section,  or 


Thus,  for  a  heavy  10-inch  beam  it  is  found  by  the  table 
in  Art.  33  that  r2  =  9.50/1  1.8  =  0.805  inches2. 

The  values  of  r2  for  rectangles  of  least  side  d  are  readily 
obtained  from  the  moments  of  inertia  given  in  Art.  24; 
For  a  solid  rectangle,     r2  =%2cP 

fcc^-fcidi3 

For  a  hollow  rectangle,  rz==     /L  ,  —  r-  j^- 
12(bd—bidi) 

For  a  solid  square,         i 
For  a  hollow  square,      r2 

The  reason  why  the  least  moment  of  inertia  is  used 
for  columns  is  that  the  bending  tends  to  occur  in  a  plane 
perpendicular  to  the  axis  about  which  the  moment  of 
inertia  is  the  least.  Thus,  a  rectangular  strut  bends  in  a 
plane  parallel  to  the  least  side  of  its  cross-section. 

As  circular  cross-sections  are  frequently  used  for 
columns,  the  values  of  the  moment  of  inertia  for  these 
will  here  be  stated.  Let  d  be  the  outer  diameter  and  di 
the  inner  diameter.  Then 


For  a  solid  circle,     /  =  Vo^ird4,  rz  = 

For  a  hollow  circle,  /  =  %4*-(d4  -  dfi  ,    r2  =  V\  e(d2+dl2) 


80  COLUMNS   OR   STRUTS  CH.  5 

Here  the  values  of  r2  are  found  by  dividing  the  first  / 
by  %Trd2  and  the  second  7  by  %Tr(d2  —  d-f],  these  being 
the  areas  of  the  cross-sections. 

From  the  last  paragraph  in  Art.  24  it  is  seen  that  r2 
is  the  average  of  all  the  values  of  z2  for  the  cross-section. 
There  is,  however,  no  way  of  finding  this  average  r2 
except  by  first  determining  7  and  then  dividing  it  by  the 
area  A. 

Prob.  374.  Compute  the  radius  of  gyration  for  a  circular  ring 
of  10  inches  outer  and  8  inches  inner  diameter. 

Prob.  37  B.  Find  the  least  radius  of  gyration  for  a  standard 
angle  4X4  inches  and  weighing  18.5  pounds  per  linear  foot  (see 
American  Civil  Engineers'  Pocket  Book,  Sect.  4,  Art.  48). 

ART.  38.     FORMULA  FOR  COLUMNS 

Columns  and  struts  generally  fail  under  the  stresses 
produced  by  combined  compression  and  bending.  The 
phenomena  are  so  complex  that  no  purely  theoretical 
formula  will  fully  represent  all  cases.  The  formula  of 
Rankine  is  that  which  has  the  best  rational  basis,  but 
this  cannot  here  be  fully  developed  as  the  laws  of  deflection 
have  not  yet  been  discussed. 

Let  P  be  the  load  on  the  vertical  column,  and  let  a 
horizontal  plane  ab  cut  it  at  the  middle  (Fig.  31).  If  A  is 
the  section  area,  the  average  compressive  unit-stress  P/A 
may  be  represented  by  the  line  cd.  But  in  consequence 
of  the  bending  this  is  increased  to  aq  on  the  concave  side 
and  decreased  to  bq  on  the  convex  side.  The  triangles 
pdq  and  qdp  represent  the  longitudinal  bending  stresses, 
as  in  beams.  Let  the  maximum  unit-stress  aq  be  denoted 
by  S.  The  part  ap  is  equal  to  cd  or  to  P/A.  The  part 


ART.  38 


FORMULA    FOR   COLUMNS 


81 


pq  is  due  to  the  bending  and  will  be  denoted  by  Si.  Hence 
the  maximum  unit-stress  aq  is  given  by 

S-f+A 

Now  from  the  formula  (4)  established  for  cases  of  bending 
in  Art.  30,  the  value  of  Si  is  Me/ 1,  where  M  is  the  bending 


Fig.  31 


Fig.  32 


moment  of  the  external  forces.     Here  the  only  external 
force  is  P,  and  its  lever  arm  is  the  lateral  deflection  of 
the  central  line  of  the  column.    Let  this  lateral  deflection 
be  called  /;  then  M  =  Pf,  and  accordingly, 
^  =  P     PC/ 

where  c  represents  the  distance  ac  in  the  figure. 

Now  let  /  be  replaced  by  Ar2,  where  r  is  the  radius  of 
gyration  of  the  cross-section.  Then  the  preceding  equation 
becomes 


82 


COLUMNS   OR   STRUTS 


CH.  5 


which  shows  how  the  unit-stress  S  on  the  concave  side 
increases  with  the  lateral  deflection  /.  By  a  discussion 
of  the  subject  of  deflection,  such  as  is  given  in  'Mechanics 
of  Materials/  it  is  shown  that  the  value  of  /  which  is 
liable  to  occur  increases  as  the  square  of  the  length  of  the 
beam,  or  cf  may  be  made  equal  to  ql2,  where  q  depends 
upon  the  kind  of  material  and  the  arrangement  of  the 
ends  (see  Art.  56).  The  last  equation  may  now  be  written, 
P  S 


which  is  Rankine's  formula  for  columns. 

The  values  of  q  to  be  used  in  problems  and  examples 
in  this  book  are  given  in  the  following  table.  These 
mean  values  have  been  derived  by  the  consideration  of 
numerous  experiments  on  the  rupture  of  columns  and 

TABLE  9.     COLUMN  CONSTANTS  q 


Material 

Both  Ends 
Fixed 

One  Fixed  End 
and 
One  Round  End 

Both  Ends 
Round 

Timber 

1 

3000 

2 
3000 

4 

3000 

Cast  Iron 

1 
5000 

2 

5000 

4 
5000 

Wrought  Iron 

1 

2 

4 

35000 

35  000 

35000 

Steel 

1 
25000 

2 

4 

25000 

25000 

struts.  It  is  seen  that  in  all  cases  q  is  four  times  as  large 
for  round  ends  as  for  fixed  ends,  which  results  from  the 
fact  that  a  very  long  column  with  round  ends  has  only 
one-fourth  the  strength  of  one  with  fixed  ends. 


ART.  39     SAFE  LOADS  FOR  COLUMNS         83 

Prob.  38  A.  If  P/A=5QQ  pounds  per  square  inch  for  a  timber 
column  with  fixed  ends,  find  from  formula  (5)  the  values  of  S  when 
l/r=Q,  l/r  =  50,  and  l/r  =  lQO. 

Prob.  38  B.  When  the  length  I  becomes  very  small,  show  that 
formula  (5)  reduces  to  formula  (1). 

ART.  39.    SAFE  LOADS  FOR  COLUMNS 

To  find  a  safe  load  for  a  column  of  given  size  and 
material  the  working  value  of  S  is  to  be  assumed  by 
Art.  7.  The  value  of  r  is  determined  by  Art.  37,  and  q 
by  the  table  in  Art.  38.  Then,  from  the  formula  (5), 

AS 


which  gives  the  safe  load  P  for  the  column. 

For  example,  let  it  be  required  to  find  the  safe  load  for 
a  timber  strut  3X4  inches  and  5  feet  long,  so  that  the 
greatest  compressive  unit-stress  S  may  be  800  pounds 
per  square  inch.  Here  6  =  4  inches,  d  =  3  inches,  r2  = 
Vi2d2  =  M  inches2,  Z2  =  3600  inches2,  Z2/r2  =  4800,  q  =  1/3000, 
ql*/r2=l.Q.  Then 

D     12X800     , 

P  =  =  3  690  pounds 

which  is  the  safe  load  for  the  strut.  If  the  length  is 
only  about  one  foot,  the  safe  load  will  be  simply  P  = 
12X800  =  9600  pounds.  If  the  length  is  12  feet,  P  will 
be  found  by  the  formula  to  be  only  940  pounds.  The 
influence  of  the  length  on  the  safe  load  is  hence  very  great. 

Prob.  39  A.  A  hollow  cast-iron  column  to  be  used  in  a  building 
is  6X6  inches  outside  dimensions  and  5X5  inches  inside  dimen- 
sions, the  length  being  18  feet,  and  the  ends  fixed.  Find  its  safe  load. 


84  COLUMNS   OR   STRUTS  CH.  5 

Prob.  39  B.  Find  the  safe  load  for  the  piston-rod  of  a  steam 
engine,  its  diameter  being  2  inches  and  its  length  36  inches,  when  the 
allowable  value  of  S  is  5000  pounds  per  square  inch. 

ART.  40.     INVESTIGATION  OF  COLUMNS 

The  investigation  of  a  column  under  a  given  load 
consists  in  computing  the  unit-stress  S  from  formula  (5) 
and  then  comparing  this  with  the  ultimate  strength  and 
elastic  limit  of  the  material,  having  due  regard  to  whether 
the  stresses  are  steady,  variable,  or  sudden  (Art.  7).  The 
value  of  S  is 


and  the  given  data  will  include  all  the  quantities  in  the 
second  member. 

For  example,  a  wrought-iron  tube  used  as  a  column 
with  fixed  ends  carries  a  load  of  38  000  pounds.  Its 
outside  diameter  is  ,6.36  inches,  its  inside  diameter  6.02 
inches,  and  its  length  18  feet.  It  is  required  to  find  the 
unit-stress  S  and  the  factor  of  safety.  Here  P  =  38  000 
pounds,  J.  =  K7r(6.362-6.022)=3.31  square  inches,  q  = 
1/35000,  2=18X12  =  216  inches,  r2  =  Vi6(6.362+6.022)  = 
4.79  inches2.  Then  by  the  formula 

q     38000/j        216X216  _\ 
3.31   V  +35000X4.79/ 

or  £=14700  pounds  per  square  inch.  The  factor  of 
safety  is  thus  about  4,  which  is  a  safe  value  if  the  col- 
umn is  used  under  steady  stress,  but  too  small  if  sudden 
stresses  or  shocks  are  liable  to  occur.  If  the  length  of  this 
column  is  36  feet,  the  unit-stress  S  will  become  about 
25  000  pounds  per  square  inch,  so  that  its  factor  of  safety 
will  be  only  2.2,  a  value  far  too  low  for  proper  security. 


ART.  41  DESIGN  OF  COLUMNS  85 

As  a  second  example  let  a  heavy  10-inch  steel  I  beam 
which  is  25  feet  long  be  used  as  a  strut  in  a  bridge  truss, 
the  ends  being  hinged  on  pins.  Let  the  compression 
on  it  be  5900  pounds.  Here  from  the  table  in  Art.  33 
there  is  found  A  =  11.8  square  inches  and  /'  =  9.50  inches4, 
whence  r2  =  0.80  inches2;  also  q  =  4/25  000,  Z=300  inches, 
P  =  5900  pounds.  Then,  from  the  formula,  S  is  found 
to  be  9500  pounds  per  square  inch,  which  is  about  one- 
third  of  the  elastic  limit  of  the  material,  and  hence  a 
safe  value. 

Prob.  404.  A  pine  stick  3X3  inches  and  12  feet  long  is  used 
in  a  building  as  a  column  with  fixed  ends.  Find  its  factor  of  safety 
under  a  load  of  3000  pounds.  If  its  length  is  only  one  foot,  what  is 
the  factor  of  safety? 

Prob.  40  B.  A  rectangular  wooden  column,  12X12  inches  in 
outside  dimensions  and  9X9  inches  in  inside  dimensions,  is  14  feet 
long.  Compute  the  unit-stress  S  when  the  load  P  is  10  000  pounds 
and  the  ends  are  fixed. 

ART.  41.     DESIGN  OF  COLUMNS 

When  the  length  of  a  column  is  given  and  the  load  to  be 
carried  by  it,  the  design  consists  in  selecting  the  proper 
material  and  then  finding  the  dimensions  so  that  the  unit- 
stress  S  in  formula  (5)  may  have  the  proper  value.  This 
is  often  done  by  trial,  dimensions  being  assumed  and 
inserted  in  (5),  and  if  these  do  not  fit,  changes  are  made 
in  them  until  a  satisfactory  agreement  is  found.  For 
example,  let  it  be  required  to  find  the  size  of  a  square 
wooden  column  with  fixed  ends  and  24  feet  long  to  carry 
a  load  of  100  000  pounds  with  a  unit-stress  S  of  800 
pounds  per  square  inch.  If  the  column  is  very  short  the 
area  A  should  be  100000/800=125  square  inches,  and 
the  side  of  the  square  about  11  inches.  The  column 


86  COLUMNS    OR   STRUTS  CH.    5 

24  feet  long  must  be  larger  than  this;  assume  it  16  inches. 
Then,  from  the  formula  of  the  last  article  find  the  value 
of  S;  this  being  a  little  larger  than  800  shows  that  16 
inches  is  too  small.  Again,  trying  17  inches,  S  is  found 
to  be  a  little  smaller  than  800.  Hence  16  3^  inches  is  an 
approximate  solution  of  the  problem. 

Equations  can  be  derived,  however,  for  finding  the 
size  of  solid  square  and  round  columns  by  placing  for  A 
and  r2  in  formula  (5)  their  values  in  terms  of  the  side 
or  diameter  d.  Thus  for  a  solid  square  column 

P       P 

d*-^d2  =  ^. 

o         o 

and  for  a  solid  round  column 


7TO  7T 

As  an  example,  take  the  data  of  the  last  paragraph,  where 
P=100000,  S  =  SOO,  g=l/3000,  and  Z  =  24X12.  Insert- 
ing these  in  the  first  equation  it  becomes  d4—  125d2  = 
41  472,  and  solving,  there  is  found  d2  =  275.5,  whence  d= 
16.6  inches  is  the  side  of  the  square  column. 

For  hollow  square  and  round  columns  equations  can 
be  derived  in  a  similar  way  for  finding  the  inner  side  or 
diameter  dt  when  the  outer  side  or  diameter  d  is  given. 
Thus  for  a  hollow  square  column 


and  for  a  hollow  round  column 


For  example,  let  it  be  required  to  find  the  inner  diameter 


ART.  42 


ECCENTRIC   LOADS 


87 


d\  for  a  cast-iron  hollow  round  column  with  fixed  ends, 
which  is  18  feet  long  and  10  inches  outer  diameter,  and 
which  is  to  carry  a  steady  load  of  240  000  pounds.  Here 
the  working  value  of  S  is  15  000  pounds  per  square  inch, 
and  q=  1/5000.  Then  the  last  equation  gives  di2  =  Q0.7 
whence  ^  =  7.8  inches  for  the  inner  diameter. 

Prob.  41  A.  Find  what  steel  I  beam  12  feet  long  may  be  used 
as  a  column  to  carry  a  load  of  100  000  pounds,  taking  the  working 
value  of  S  at  12  000  pounds  per  square  inch. 

Prob.  41  B.  A  hollow  square  column  of  wood  with  fixed  ends 
and  14  feet  long  has  outside  dimensions  of  12X12  inches  and  carries 
a  load  of  9450  pounds.  Find  the  inside  dimensions  so  that  S  shall 
be  900  pounds  per  square  inch. 

ART.  42.     ECCENTRIC  LOADS 

Thus  far  it  has  been  supposed  that  the  load  is  so  applied 
to  the  end  of  a  column  that  its  line  of  action  coincides 


C 

Fig.  33 

with  the  axis  of  the  column.  In  many  instances,  however, 
this  is  not  the  case.  Let  Fig.  33  represent  a  short  block 
where  the  load  P  is  applied  at  a  distance  e  from  the  axis 


88  COLUMNS    OR   STRUTS  CH.  5 

passing  through  the  center  of  gravity  of  the  cross-section. 
The  distribution  of  the  internal  compressive  unit-stresses 
in  every  section  is  then  not  uniform.  The  mean  unit- 
stress  on  the  area  A  is  P/A,  but  this  is  increased  on  -the 
side  nearest  P  and  decreased  on  the  opposite  side  by  the 
unit-stress  due  to  the  flexure.  Let  CC  be  the  neutral 
axis  of  the  cross-section  and  c  the  distance  to  the  side, 
let  /  be  the  moment  of  inertia  and  r  the  radius  of  gyration 
of  the  cross-section  with  respect  to  the  axis  (7(7;  let  S' 
be  the  flexural  unit-stress  at  the  side  of  the  column.  Then 
from  the  flexure  formula  (4),  Sf  =  Me/  1.  But  the  bending 
moment  M  is  Pe,  hence  S'  =  Pec/I  =  Pec/Ar2.  Adding 
this  to  the  mean  unit-stress  P/A,  there  results 


which  is  the  compressive  unit-stress  on  that  side  of  the 
column  nearest  P.  On  the  other  side  of  the  column 
the  unit-stress  is  found  by  changing  the  +  sign  to  —  . 
A  small  eccentricity  e  causes  the  unit-stress  S  to  deviate 
much  from  the  mean  value  P/A.  For  a  rectangular 
section  r2  =  Mad2  and  c  =  %d,  so  that 

P  f         &\ 
For  side  A  of  the  prism,     <Si  =  -r  (  1+6-5  I 

A.  \  Gil 

For  side  B  of  the  prism,     Sz=-r(l—6-?} 

A.  \         d/ 

When  e  =  y6d,  then  Si  =  2P/A  which  is  double  the  mean 
value,  while  £2  =  0.  When  e  =  yad,  then  Si  =  3P/A  and 
£2=  —P/A;  hence  the  side  B  is  under  tension  instead  of 
compression.  It  is  thus  seen  that,  in  placing  loads  on  a 
column,  eccentricity  of  application  should  be  avoided. 

The  above  formula  (5)    applies  to  a  short  column  or 


ART.  43          THE  STRAIGHT-LINE  FORMULA  89 

to  one  in  which  l/r  does  not  exceed  10.  For  longer  col- 
umns it  is  customary  to  add  the  quantity  ce/r2  to  the 
denominator  in  Rankine's  formula,  which  thus  becomes 

?- ^— :  (»>' 


This  formula  may  be  used  for  finding  the  safe  load  on 
a  column  having  an  eccentric  load,  for  investigating  an 
existing  column,  or  for  designing  a  section  for  a  proposed 
column. 

Prob.  42  A.  Using  formula  (5)'  find  the  safe  load  for  the  data 
given  in  Prob.  39  A,  taking  the  eccentricity  of  the  load  as  %  inches. 

Prob.  42  B.  Using  formula  (5)'  find  the  factor  of  safety  for  the 
data  given  in  Prob.  40  A,  taking  the  eccentricity  of  the  load  as 
%  inches. 

ART.  43.     THE  STRAIGHT-LINE  FORMULA 

Another  formula  for  columns  is  that  called  the  straight- 
line  formula,  because  the  relation  between  P/A  and  l/r 
is  the  same  as  that  between  y  and  x  in  the  equation  of  a 
straight  line.  This  formula  is 

f-s-d 

A  r 

in  which  S  is  the  unit-stress  on  the  concave  side  of  the 
column  and  C  is  a  quantity  which  varies  with  the  material 
and  the  condition  of  the  ends.  For  columns  with  fixed 
ends  which  are  used  in  buildings  under  steady  loads  the 
following  are  used  in  cases  of  design: 

For  cast  iron,  P/A  =  10  000 - 40(Z/r) 

For  wrought  iron,     P/A  =  1 2  000  -  60 (l/r) 
For  structural  steel,  P/A  =  16  000-70(J/r) 


90  COLUMNS   OR   STRUTS  CH.  5 

These  formulas  only  apply  when  P  is  in  pounds,  A  in 
square  inches,  and  when  the  value  of  l/r  is  less  than  120. 
They  do  not  have  the  same  degree  of  reliability  as  Ran- 
kine's  formula,  since  they  are  wholly  empirical.  When 
specifications  require  that  they  should  be  used,  this  must 
be  done,  but  otherwise  Rankine's  formula  (5)  should  be 
employed. 

For  example,  find  the  safe  load  for  a  hollow  cast-iron 
column  6X6  niches  in  outside  dimensions  and  5X5  inches 
in  inside  dimensions,  the  length  being  18  feet  and  the  ends 
fixed.  Here  A  =  11  square  inches,  r2  =  Vis(36+25)  =  5.08 
whence  r— 2. 252  inches,  Z/r  =  95.9,  and  then  from  the 
formula  P=67800  pounds.  In  this  solution  no  use  is 
made  of  the  unit-stress  S  on  the  concave  side  of  the 
column.  By  Rankine's  formula,  using  S=  15  000  pounds 
per  square  inch,  there  is  found  P=58  100  pounds,  which 
is  a  more  reliable  value. 

Again,  let  it  be  required  to  find  the  diameter  of  a  solid 
cast-iron  strut  6  feet  long  to  safely  carry  a  steady  load  of 
64  000  pounds.  Here  for  a  very  short  strut,  where  1  =  0, 
the  area  required  is  A  =  64  000/10  000  =  6.4  square  inches, 
which  corresponds  to  d  =  2.85  and  r  =  0.71  inches.  Assume 
then  d  =  4  inches,  whence  .4  =  12.57  square  inches,  r=l 
inch,  and  l/r  =  72 ;  inserting  these  in  the  formula  there  is 
found  P  =  89  000  pounds  which,  being  greater  than  the 
given  value,  shows  that  4  inches  is  too  large  a  diameter. 
Assume  again  that  d  =  3.5  inches,  whence  A  =  9.62  square 
inches,  r  =  0.875  and  Z/r  =  84.6;  inserting  these  in  the 
formula,  there  is  found  P  =  63  600  pounds,  which  is  very 
close  to  the  given  value,  so  that  d  =  3.5  inches  is  a  satisfac- 
tory solution  of  the  problem  by  the  straight-line  formula. 


ART.  44  REVIEW  PROBLEMS  91 

Prob.  43.  A  column  of  structural  steel  has  the  dimensions 
stated  in  Art.  25  for  Fig.  19.  What  steady  load  can  it  carry  accord- 
ing to  the  straight-line  formula? 

ART.  44.     REVIEW  PROBLEMS 

Prob.  44  A.  Find  the  safe  steady  load  for  a  hollow  short  cast- 
iron  column  which  is  12  inches  in  outside  and  9  inches  in  inside 
diameter. 

Prob.  44  B.  Given  g  =  %ooo  and  <S  =  9000  pounds  per  square 
inch  for  a  cast-iron  column.  Plot  a  curve  for  formula  (5),  taking 
values  of  l/r  as  abscissas  and  values  of  P/A  as  ordinates. 

Prob.  44  C.  Determine  the  safe  load  for  a  fixed-ended  timber 
column  3X4  inches  in  section  and  10  feet  long,  so  that  the  greatest 
compressive  unit-stress  may  be  800  pounds  per  square  inch. 

Prob.  44  D.  A  cylindrical  wrought-iron  column  with  fixed  ends 
is  12  feet  long,  6.36  inches  in  outside  diameter,  6.02  inches  in  inside 
diameter,  and  carries  a  load  of  49  000  pounds.  Find  its  factor  of 
safety. 

Prob.  44  E.  Compute  the  size  of  a  square  timber  column  with 
fixed  ends  to  carry  a  load  of  100  000  pounds  with  a  factor  of  safety 
of  10,  its  length  being  12  feet. 

Prob.  44  F.  A  beam  25  feet  long  carries  a  uniform  load  of  3000 
pounds  per  linear  foot,  and  is  supported  at  its  ends  by  two  round 
cast-iron  columns  15  feet  long.  The  columns  have  fixed  ends  and 
are  6  inches  in  outer  diameter.  Find  the  inner  diameter  of  the  col- 
umns so  that  the  unit-stress  S  is  10  000  pounds  per  square  inch. 

Prob.  44  G.  A  10-inch  standard  I  beam  weighing  30  pounds  per 
linear  foot  is  used  as  one  of  the  compression  members  in  a  small 
bridge.  The  column  is  fixed-ended  and  is  20  feet  long.  Will  the 
column  be  safe  for  a  load  of  50  000  pounds? 

Prob.  44  H.  The  piston-rod  of  an  engine  is  circular  in  shape  and 
its  stroke  is  three  feet.  The  maximum  load  upon  the  piston  is 
20  000  pounds.  Find  the  proper  diameter  for  the  rod,  using  S  as 
6000  pounds  per  square  inch. 


92 


THE   TORSION    OF   SHAFTS 


CH.  6 


CHAPTER  6 

THE   TORSION   OF  SHAFTS 
ART.  45.     PHENOMENA  OF  TORSION 

Torsion  is  that  kind  of  stress  which  occurs  when 
external  forces  tend  to  twist  a  body  round  an  axis.  A 
shaft  which  transmits  power  is  twisted  by  the  forces 
applied  to  the  pulleys,  and  thus  all  its  cross-sections  are 
brought  into  stress.  This  stress  is  a  kind  of  shearing 
but  the  forces  acting  in  different  parts  of  a  section  are  not 
parallel. 

Let  one  end  of  a  horizontal  bar  be  rigidly  fixed,  and 
to  the  free  end  let  a  lever  be  attached  at  right  angles 


Fig.  34 

to  its  axis  (Fig.  34).  A  weight  P  hung  at  the  end  of  this 
lever  will  twist  the  shaft  so  that  a  line  ab  which  originally 
was  horizontal  will -assume  a  spiral  form  ad,  while  the 
radial  line  cb  will  move  to  the  position  cd.  It  has  been 
shown  by  experiments  that,  if  the  material  is  not  stressed 
beyond  its  elastic  limit,  the  angles  bed  and  bad  are  pro- 


ART.  45  PHENOMENA  OF  TORSION  93 

portional  to  the  applied  weight  P,  and  that  on  the  removal 
of  this  weight  the  lines  cd  and  ad  will  return  to  their 
original  positions.  If  the  elastic  limit  is  exceeded  this 
proportionality  does  not  hold,  and  if  the  stress  is  made 
great  enough  the  bar  will  be  ruptured. 

Let  p  be  the  lever-arm  of  P  with  respect  to  the  axis  c. 
Then  experience  also  shows  that  the  amount  of  twist 
is  proportional  to  p.  The  product  Pp  is  the  moment  of  P 
with  respect  to  the  axis,  and  it  is  called  the  'twisting 
moment.'  If  there  are  several  forces  PI,  PZ,  etc.,  acting 
on  the  shaft  with  lever-arms  p\,  pz,  etc.,  the  total  twisting 
moment  Pp  is  the  algebraic  sum  of  the  separate  moments 
Pipi,  P^pz,  etc.,  those  being  positive  which  tend  to  turn 
in  the  direction  of  the  hands  of  a  watch,  and  those  negative 
which  turn  in  the  opposite  direction. 

For  example,  let  the  three  lever-arms  be  applied  to 
a  bar  at  the  points  B,  C,  and  D,  whose  distances  from  A 


Fig.  35 

are  5,  8,  and  12  feet.  Let  the  forces  in  Fig.  35  be  PI  =  30 
pounds,  P2  =  60  pounds,  and  P3— 100  pounds,  their  lever- 
arms  being  pi  =  2.5  feet,  p2  =  2.0  feet,  and  ps  =  3.5  feet. 
Then  for  all  sections  between  D  and  C  the  twisting 
moment  is  +30X2.5= +75  pound-feet;  for  all  sections 
between  C  and  B  the  twisting  moment  is  +30X2.5  — 


94  THE   TORSION   OF  SHAFTS  CH.  6 

60X2.0=— 45  pound-feet;  and  for  all  sections  between 
B  and  A  the  twisting  moment  is  +30X2.5  —  60X2.0+ 
100  X  3.5  =  +305  pound-feet.  Thus  the  tendency  to  twist 
between  B  and  C  is  in  the  opposite  direction  to  that  in 
the  other  parts  of  the  bar. 

Prob.  45  A.  If  a  force  of  600  pounds  acting  at  5  inches  from  the 
axis  twists  the  end  of  a  shaft  30  degrees,  what  force  acting  at  12 
inches  from  the  axis  will  twist  it  60  degrees? 

Prob.  45  B.  It  is  found  by  experiment  that  the  angle  bed  in 
Fig.  34  is  proportional  to  the  length  of  the  bar  when  P  and  p  are 
constant.  If  the  angle  bed  is  6°  35'  for  a  shaft  9.4  feet  long,  what  will 
this  angle  be  for  a  shaft  13.5  feet  long? 

ART.  46.    POLAR  MOMENTS  OF  INERTIA 

In  the  discussion  of  shafts  the  moments  of  inertia  of 
cross-sections  are  required  with  respect  to  a  point  at  the 
center  of  the  shaft  and  not  with  respect  to  an  axis  in  the 
same  plane,  as  in  beams  and  columns.  The  'polar 
moment  of  inertia '  of  a  surface  is  denned  as  the  sum  of  the 
products  obtained  by  multiplying  each  elementary  area 
by  the  square  of  its  distance  from  the  center  of  the  surface. 
Thus  if  a  be  any  elementary  area  and  x  its  distance  from 
the  center  the  quantity  Sao;2  is  the  polar  moment  of  inertia, 
2  being  the  symbol  of  summation,  which  denotes  that 
all  the  values  a^Xi,  a2x?,  etc.,  are  to  added  until  the 
entire  surface  is  covered. 

In  Fig.  36  let  a  be  any  elementary  area  and  z  its  dis- 
tance from  an  axis  AB  passing  through  the  center  of 
gravity  of  the  section;  then  2a22,  or  the  summation  of 
all  the  values  of  az2,  is  the  moment  of  inertia  with  respect 
to  the  axis  AB  (Art.  24).  Also,  if  y  is  the  distance  from 
a  to  an  axis  CD  which  is  normal  to  A  B,  then  Sa?/2  is  the 


ART.  46 


POLAR   MOMENTS   OF   INERTIA 


95 


moment  of  inertia  with  respect  to  the  axis  CD.    But  since 

22 -f- ?/2  =  s;2,    the   product    Sax2   is   equal   to    2az2-}-2a?/2; 

that  is,  the  polar  moment  of  inertia  is  the  sum  of  the 

D 


-&• 


Fig.  36 

moments  of  inertia  taken  with  respect  to  any  two  rec- 
tangular axes. 

The  polar  moment  of  inertia  is  represented  by  J. 
By  the  aid  of  the  above  principle  its  value  is  readily 
found  from  the  values  of  /  given  in  Arts.  24  and  37.  Let 
d  be  the  diameter  of  a  circle;  then, 

For  a  solid  circle,      J=y32Trd* 

Also,  in  the  case  of  a  hollow  section,  let  d  be  the  outer 
and  di  be  the  inner  diameter;  then, 

For  a  hollow  circle,      J  =  1/3271- (d4 — di4) 

The  circular  sections  are  most  frequently  used  for  shafts, 
and  the  discussions  of  this  chapter  apply  mainly  to  such 
shafts.  The  theory  of  the  torsion  of  square  and  rec- 
tangular bars  is  very  complicated  and  cannot  be  given  here. 

Prob.  46  A.  Show  that  the  polar  moment  of  inertia  for  the  hollow 
circular  section  is  ]^A(d'2+di2),  where  A  is  the  section  area. 

Prob.  46  B.  Show  that  the  polar  moment  of  inertia  for  a  square 
section  area  is  M)d4. 


96 


THE   TORSION    OF   SHAFTS 


CH.  G 


ART.  47.    FORMULA  FOR  TORSION 

If  two  cross-sections  are  taken  in  a  shaft  very  near 
together,  each  section  tends  to  twist  with  respect  to  the 
other,  and  shearing  stresses  are  found  to  exist  in  all  parts 
of  the  section.  These  stresses  are  zero  at  the  center  and 
greatest  at  the  boundary  of  a  circular  section,  and  they 
act  everywhere  perpendicular  to  the  lever-arms  drawn 
to  them  from  the  center.  If  the  elastic  limit  is  not 
exceeded  it  is  found  that  the  stresses  are  proportional  to 
their  lever-arms. 

Let  P  be  the  force  acting  with  the  lever-arm  p  which 
produces  the  twisting  moment  Pp  (Fig.  37).  This  must 


Fig.  37 

be  equal  to  the  resisting  moment  of  the  internal  stresses. 
Let  S  be  the  shearing  unit-stress  at  the  remotest  part  of 
the  section  whose  distance  from  the  center  is  c.  Then  the 
stress  at  a  distance  Y^c  from  the  center  is  %S,  and  the 
stress  at  a  distance  x  from  the  center  is  Sx/c.  The  total 
stress  on  an  elementary  area  a  at  a  distance  x  from  the 
center  is  then  aSx/c,  and  the  moment  of  this  stress  with 
respect  to  the  center  is  (S/c)ax2.  The  resisting  moment  is 
the  sum  of  all  the  values  of  S/cax2,  or,  since  S  and  c  are 
constants,  this  sum  is  GS/c)  2az2.  But,  as  seen  in  the  last 


ART.  47  FORMULA  FOR  TORSION  97 

article,  the  quantity  Sax2  is  the  polar  moment  of  inertia  J. 
Accordingly  the  resisting  moment  of  the  internal  shearing 
stresses  is  SJ/c,  and,  equating  this  to  the  twisting  moment 
Pp,  there  results 

S-^=Pp  (6) 

which  is  the  fundamental  formula  for  the  torsion  of 
shafts  with  circular  cross-sections. 

This  formula  is  analogous  with  formula  (4)  for  beams, 
and  is  used  in  a  similar  manner  to  investigate  and  design 
shafts.  The  unit-stress  S  is  here  always  a  shearing  stress, 
and  its  working  values  are  to  be  determined  by  applying 
factors  of  safety  to  the  ultimate  shearing  strengths  given 
in  Art.  6.  Shafts  which  transmit  power  are  subject  to 
variable  loads,  and  often  to  shocks,  and  hence  their  values 
of  S  should  be  taken  low.  Formula  (6)  is  subject  to  the 
same  limitation  as  formula  (4),  namely,  it  is  only  true 
when  the  unit-stress  S  is  less  than  the  elastic  limit  of  the 
material  (see  Art.  66). 

For  example,  the  twisting  moment  Pp  being  20  000 
pound-inches,  it  is  required  to  find  the  shearing  stress  pro- 
duced by  it  in  a  circular  shaft  4  inches  in  diameter.  Here 
c  =  2  inches,  ,7  =  25.13  inches4,  and  then  by  (6)  the  value  of 
S  is  found  to  be  1590  pounds  per  square  inch.  If  the 
shaft  is  of  wood  this  is  too  low  a  value  of  S,  it  being 
about  one-half  of  the  ultimate  shearing  strength;  if  it 
is  of  wrought  iron  or  steel  there  is  a  high  factor  of  safety. 

Prob.  47  A.  A  round  steel  shaft  is  subject  to  a  twisting  moment 
of  2500  pound-inches.  What  should  be  its  diameter  so  that  the 
greatest  shear  S  may  be  6000  pounds  per  square  inch? 

Prob.  47  B.    A  pulley  36  inches  in  diameter  is  placed  on  a  2-inch 

7 


98  THE   TORSION    OF   SHAFTS  CH.  6 

wrought-iron  shaft,  and  the  effective  pull  of  the  belt  on  the  pulley 
is  500  pounds.    What  is  the  factor  of  safety  of  the  shaft? 

ART.  48.    SHAFTS  TO  TRANSMIT  POWER 

'  Work '  is  the  product  of  a  force  by  the  distance  through 
which  it  is  exerted.  Thus,  if  a  weight  of  10  pounds  is 
lifted  vertically  a  distance  of  5  feet  there  are  performed 
50  foot-pounds  of  work.  If  this  weight  is  moved  hori- 
zontally, however,  the  force  required  depends  only  on 
frictional  and  other  resistances;  if  these  require  a  force 
of  3  pounds  and  this  be  exerted  through  a  distance  of 
5  feet,  then  15  foot-pounds  of  work  are  performed. 

'Power'  is  work  performed  in  a  given  time.  The  unit 
of  power  is  the  'horse-power,'  which  is  denned  as  33  000 
foot-pounds  of  work  performed  in  one  minute.  Thus,  if 

99  000  foot-pounds  of  work  are  performed  in  one  minute, 
the  power  exerted   is  3   horse-powers;    if   99  000   foot- 
pounds of  work  are  performed  in  two  minutes,  the  power 
exerted  is  1^  horse-powers. 

Power  from  a  motor  is  usually  transmitted  to  a  shaft 
by  belts,  and  the  shaft  then  transmits  the  power  to  the 
places  where  the  work  is  to  be  performed.  In  doing 
this  the  shaft  is  brought  under  stress.  Let  H  be  the 
power  transmitted  through  a  belt  to  a  pulley.  Let  P 
be  the  tangential  force  in  pounds  brought  by  the  belt 
on  the  circumference  of  the  pulley,  and  let  p  be  the  radius 
of  the  pulley  in  inches.  Let  n  be  the  number  of  revolu- 
tions made  by  the  shaft  and  pulley  in  one  minute.  In  one 
revolution  a  force  of  P  pounds  acts  through  2irp  inches,  and 
the  work  of  PX^wp  pound-inches,  or  ^irPp  pound-feet, 
is  performed.  In  one  minute  the  work  performed  is 


ART.  49  SOLID  SHAFTS  99 


pound-feet.    The  number  of  horse-powers  exerted 
is  found  by  dividing  this  work  by  33  000,  or 
H=   nirPp 
198  000 

The  twisting  moment  Pp  may  now  be  replaced  by  the 
resisting  moment  SJ/c,  and  hence 
&/=19800Qg 
c  mr 

which  is  the  formula  for  the  discussion  of  round  shafts 
that  are  used  to  transmit  power. 

For  such  a  shaft  c  is  equal  to  one-half  of  the  outer 
diameter,  whether  its  section  be  solid  or  hollow.  The 
unit-stress  S  is  here  always  that  for  shearing,  and  in 
selecting  its  safe  value  a  high  factor  of  safety  is  to  be 
used,  as  the  shaft  is  subject  to  variable  stresses.  It  is 
noticed  that  S  varies  inversely  with  n,  that  is,  for  a 
given  power  transmitted  the  slower  the  speed  the  greater 
is  the  stress  in  the  shaft. 

Prob.  48  A  If  a  round  shaft  one  inch  in  diameter  transmits 
one  horse-power  at  100  revolutions  per  minute,  show  that  the 
shearing  stress  produced  is  about  3200  pounds  per  square  inch. 

Prob.  48  B.  A  steel  shaft  making  300  revolutions  per  minute 
is  3  inches  in  diameter.  What  horse-power  is  being  transmitted 
when  the  shearing  unit-stress  is  6000  pounds  per  square  inch? 

ART.  49.     SOLID  SHAFTS 

For  round  solid  shafts  of  diameter  d,  the  polar  moment 
of  inertia  is  V^d4,  the  value  of  c  is  }^d,  and  formula  (7) 
then  reduces  to 

£cP  =  321000-' 
n 

in  which  d  must  be  taken  in  inches  and  S  in  pounds  per 


100  THE   TORSION   OF   SHAFTS  CH.  6 

square  inch.  From  this  formula  S  may  be  found  for  a 
given  shaft  which  transmits  power,  or  d  may  be  computed 
when  it  is  required  to  design  a  shaft  for  that  purpose. 

For  example,  let  it  be  required  to  find  the  factor  of 
safety  of  a  round  solid  shaft  of  wrought  iron,  2%  inches 
in  diameter,  when  transmitting  25  horse-power  at  100 
revolutions  per  minute.  Here  d  =  2.5  inches,  #  =  25, 
n  =  100,  and  the  formula  gives 

„     321000X25 

"  =   o  £3  \7T7>rr  =  *  140  pounds  per  square  inch 
Z.o  X  1(JU 

so  that  the  factor  of  safety  is  about  10;  this  is  a  high 
value  for  a  shaft  not  subject  to  shocks. 

As  an  example  of  design,  let  it  be  required  to  find 
the  diameter  of  a  wrought-iron  shaft  when  transmitting 
90  horse-power  at  250  revolutions  per  minute.  Here 
the  factor  of  safety  will  be  taken  at  8,  or  the  allowable 
unit-stress  S  at  7000  pounds  per  square  inch.  Then, 
from  the  formula, 

,,     321  000X90 
d=  -7000X260  =16'5 
and  hence  the  diameter  d  should  be  2  %  inches. 

The  above  formulas  do  not  apply  to  square  shafts; 
in  these  the  greatest  stress  is  not  at  the  corners  but  along 
the  middle  of  the  sides.  It  is  shown  in  Mechanics  of 
Materials  (tenth  edition)  that  the  formula  for  a  solid 
square  shaft  of  side  d  is 

Sd3  =  283  600- 
n 

For  example,  let  it  be  required  to  find  how  many  horse- 
powers are  transmitted  by  a  wooden  shaft  12  inches 


ART.  50  HOLLOW  SHAFTS  101 

square  when  it  makes  25  revolutions  per  minute  and  S 
is  200  pounds  per  square  inch.  Here  all  quantities  in 
the  formula  are  known  except  H,  and  the  solution  gives 
H  =  305  horse-powers. 

Prob.  49  A.  Find  the  horse-power  that  can  be  transmitted  by  a 
solid  round  steel  shaft  of  6J^  inches  diameter  when  making  150 
revolutions  per  minute,  S  being  7500  pounds  per  square  inch. 

Prob.  49  B.  Compare  the  horse-powers  per  pound  weight  of 
shaft  which  can  be  transmitted  by  a  round  shaft  of  diameter  d 
and  a  square  shaft  of  side  d. 

ART.  50.     HOLLOW  SHAFTS 

Hollow  forged  steel  shafts  are  now  much  used  for 
ocean  steamers,  as  their  strength  is  greater  than  solid 
shafts  of  the  same  area  of  cross-section.  If  d  is  the 
outside  and  d:  the  inside  diameter,  the  value  of  J  is 
l/32ir(d4— c?i4)  and  c  is  Y^d,  These  inserted  in  (7)  give 

S   d"~dl'  =321000- 
d  n 

which  is  the  formula  for  investigation  and  discussion. 

For  example,  a  nickel  steel  shaft  of  17  inches  outside 
diameter  is  to  transmit  16  000  horse-powers  at  50  revolu- 
tions per  minute;  what  should  be  the  inside  diameter 
so  that  the  unit-stress  S  may  be  25  000  pounds  per  square 
inch?  Here  everything  is  given  except  di,  and  from  the 
equation  its  value  is  found  to  be  11  inches  nearly.  The 
area  of  the  cross-section  of  this  shaft  will  be  about  132 
square  inches,  and  its  weight  per  linear  foot  about  449 
pounds. 

Prob.  50^4.  If  a  hollow  shaft  has  the  same  area  of  cross-section 
as  a  solid  one,  and  if  the  inside  diameter  of  the  hollow  shaft  is 


102  THE   TORSION   OF   SHAFTS  CH.  6 

one-half  of  the  outside  diameter,  prove  that  the  hollow  shaft  is  44 
percent  stronger  than  the  solid  one. 

Prob.  50  B.  The  tail  shaft  of  a  marine  engine  is  15  inches  outside 
and  10  inches  inside  diameter.  What  horse-power  is  being  trans- 
mitted when  the  shaft  is  making  300  revolutions  per  minute  and  the 
unit-stress  S  is  8000  pounds  per  square  inch? 

ART.  51.     REVIEW  PROBLEMS 

/ 

Prob.  51  A.  If  a  force  of  80  pounds,  acting  at  18  inches  from 
the  axis,  twists  the  end  of  a  shaft  through  15  degrees,  what  force 
will  produce  the  same  result  when  acting  at  4  feet  from  the  axis? 

Prob.  51  B.  Compute  the  polar  moment  of  inertia  for  a  hollow 
shaft  with  outside  diameter  18  inches  and  inside  diameter  10  inches. 

Prob.  51  C.  Compute  the  shearing  unit-stress  for  the  shaft  of 
the  last  problem  when  it  is  subject  to  a  twisting  moment  of  2500 
pound-inches. 

Prob.  51  D.  Find  the  horse-power  that  can  be  transmitted  by  a 
wrought-iron  shaft  3  inches  in  diameter  when  making  50  revolutions 
per  minute,  the  value  of  S  being  6000  pounds  per  square  inch. 

Prob.  51  E.  Find  the  diameter  of  a  solid  wrought-iron  shaft  to 
transmit  90  horse-power  at  250  revolutions  per  minute,  the  value 
of  S  being  7000  pounds  per  square  inch. 

Prob.  51  F.  Find  the  ratio  of  the  strength  of  a  hollow  shaft 
to  that  of  a  solid  one,  the  section  areas  being  equal,  and  the  outside, 
diameter  of  the  hollow  section  being  three  times  as  great  as  the 
inside  diameter. 

Prob.  51  G.  The  crank  of  an  engine  is  9  inches  long  and  the 
maximum  tangential  thrust  brought  upon  it  by  the  connecting-rod 
is  5000  pounds.  Find  the  diameter  of  a  steel  shaft  to  stand  the 
above  twisting  moment  when  the  allowable  stress  S  is  6000  pounds 
per  square  inch. 

Prob.  51  H.  What  horse-power  will  be  transmitted  by  a  hollow 
shaft  of  8  inches  outside  and  5  inches  inside  diameter  when  running 
at  300  revolutions  per  minute,  the  value  of  S  being  7000  pounds  per 
square  inch?  Find  the  diameter  of  a  solid  steel  shaft  to  transmit 
the  same  horse-power  with  the  same  speed  and  unit-stress. 


ART.  52          THE  MODULUS  OF  ELASTICITY  103 

CHAPTER  7 

ELASTIC   DEFORMATIONS 
ART.  52.     THE  MODULUS  OF  ELASTICITY 

It  was  explained  in  Chapter  1  that,  when  a  bar  is 
subject  to  stresses  produced  by  gradually  applied  forces, 
the  elongations  increase  proportionately  with  the  stresses, 
if  the  elastic  limit  is  not  exceeded.  This  law  of  elasticity 
enables  the  elongations  of  bars  and  the  deflections  of 
beams  to  be  computed,  provided  none  of  the  stresses 
exceeds  the  elastic  limit  of  the  material. 

The  'modulus  of  elasticity'  in  tension  is  the  ratio 
of  the  unit-stress  to  the  unit-elongation.  Thus,  if  a  bar 
one  inch  long  and  one  square  inch  in  cross-section  is 
under  the  stress  S  an  elongation  s  is  produced,  and 

*-f  (8) 

is  the  modulus  of  elasticity.  If  the  bar  has  a  section  area 
A  which  is  acted  on  by  the  pull  P,  then  the  unit-stress  S 
is  P/A ;  if  the  bar  has  the  length  I,  an  elongation  e  is 
produced  and  the  unit-elongation  s  is  given  by  e/L 

For  compression  E  is  the  ratio  of  the  unit-stress  to 
the  unit-shortening  accompanying  that  stress,  and  in 
general  E  is  the  ratio  of  the  unit-stress  to  the  unit-deforma- 
tion. Since  s  is  an  abstract  number,  E  is  expressed  in  the 
same  unit  as  S,  that  is,  in  pounds  per  square  inch  or  kilos 
per  square  centimeter. 

Within  the  elastic  limit  S  increases  at  the  same  rate 


104 


ELASTIC    DEFORMATIONS 


CH.  7 


as  s,  and  thus  E  is  a  constant;  beyond  the  elastic  limit 
there  is  no  proper  modulus  of  elasticity.  For  different 
materials  under  the  same  unit-stress  S,  the  value  of  E 
increases  as  s  decreases;  thus  E  is  a  measure  of  the  stiff- 
ness of  materials.  Formula  (8)  also  gives 
S 


which  is  the  change  of  a  unit  of  length  of  a  bar  under 
a  given  unit-stress  S. 

The  values  of  the  moduluses  of  elasticity  for  tension 

and  compression  are  practically  the  same,  and  their  mean 

values  for  the  different  materials  are  given  in  the  follow- 

ing table.    For  shear  the  moduluses  of  elasticity  are  about 

TABLE  10.     MODULUSES  OF  ELASTICITY 


Material 

Pounds  per 
Square  Inch 

Kilos  per 
Sq.  Centimeter 

Timber 

1  500  000 

105  000 

Cast  Iron 

15  000  000 

1  050  000 

Wrought  Iron 
Steel 

25  000  000 
30  000  000 

1  750  000 
2  100  000 

one-third  of  those  stated  in  the  table.  These  values 
show  that,  within  the  elastic  limit,  steel  is  the  stiffest 
of  the  four  materials,  it  being  20  percent  stiffer  than 
wrought  iron,  twice  as  stiff  as  cast  iron,  and  twenty  times 
as  stiff  as  timber.  In  other  words,  a  given  stress  less  than 
the  elastic  limit  will  elongate  a  timber  bar  twenty  times 
as  much  as  a  steel  bar,  a  cast-iron  bar  twice  as  much, 
and  a  wrought-iron  bar  20  per  cent  more. 

Prob.  52  A.  A  bar  one  inch  square  and  2  inches  long  elongates 
0.0004  inches  under  a  tension  of  5000  pounds.  Compute  the  modulus 
of  elasticity. 


ART.  53  ELONGATION  UNDER  TENSION  105 

Prob.  52  B.  When  a  steel  bar  30  feet  long  was  subjected  to  a  ten- 
sile unit-stress  of  12  000  pounds  per  square  inch,  it  elongated  about 
0.143  inches.  Compute  the  modulus  of  elasticity  of  steel. 


ART.  53.     ELONGATION  UNDER  TENSION 

Let  a  bar  whose  section  area  is  A  and  whose  length 
is  I  be  under  the  tension  P,  and  let  e  be  the  elongation 
produced.  The  unit-stress  S  is  P/A  and  the  unit-elonga- 
tion s  is  e/L  Then  the  modulus  of  elasticity  E  is 

E=S=PI 

s    Ae 
and  hence,  if  P/A  be  less  than  the  elastic  limit, 

-Pl 
'~AE 

is  the  elastic  elongation  of  the  bar  due  to  the  applied 
tension  P. 

For  example,  let  it  be  required  to  find  the  elongation 
of  a  wrought-iron  bar  30  feet  long  when  stressed  up  to 
its  elastic  limit.  Here  P/A  =  25  000  pounds  per  square 
inch,  #  =  25  000  000  pounds  per  square  inch,  and  Z  =  360 
inches.  Then  from  the  formula,  e  =  0.36  inches.  This  is 
the  elastic  elongation;  the  ultimate  elongation  will  be 
about  72  inches.  In  all  cases,  as  seen  from  the  figure  in 
Art.  4,  the  elastic  elongations  are  very  small  compared 
with  the  ultimate  elongations. 

Prob.  53  A.  A  steel  eye-bar  30  feet  long  is  1^X6  inches  in  size. 
How  much  does  it  elongate  under  a  pull  of  90  000  pounds? 

Prob.  53  B.  What  is  the  ultimate  elongation  of  a  steel  bar  1  inch 
square  and  10  feet  long? 


106  ELASTIC   DEFORMATIONS  CH.  7 

ART.  54.    SHORTENING  UNDER  COMPRESSION 

If  a  bar  of  cross-section  A  and  length  I  be  under  the 
compression  P  it  shortens  the  amount  e.  For  a  short 
bar  where  no  lateral  deflection  can  occur  the  unit-stress 
S  is  uniform  over  the  cross-section,  and  the  shortening 
follows  the  same  law  as  does  the  elongation  in  tension, 
and  hence 

PI 
e  =  AE 

Here,  as  before,  the  unit-stress  P/A  must  not  exceed 
the  elastic  limit  of  the  material. 

For  example,  let  a  cast-iron  bar  one  inch  in  diameter 
and  5  inches  long  be  under  a  compression  of  30  000  pounds. 
Here  P  =  30000  pounds,  A  =  0.785  square  inches,  1  =  5 
inches,  and  E=I5  000  000  pounds  per  square  inch.  Then 
from  the  formula,  e  =  0.0 128  inches;  but  this  result  is  of 
no  value,  because  the  unit-stress  P/A  is  nearly  double 
the  elastic  limit  of  cast  iron.  If,  however,  P  is  given 
as  3000  pounds,  then  the  formula  properly  applies,  and  e 
is  found  to  be  0.0013  inches. 

Prob.  54  A.  A  wrought-iron  bar  18  inches  long  weighs  24  pounds. 
How  much  will  it  shorten  under  a  compression  of  7250  pounds? 

Prob.  54  B.  The  piston-rod  of  a  steam  engine  is  4  inches  in 
diameter  and  30  inches  long,  while  the  piston  is  24  inches  in  diam- 
eter. What  is  the  change  in  length  of  the  piston-rod  when  the  steam 
pressure  is  200  pounds  per  square  inch? 

ART.  55.     DEFLECTION  OF  CANTILEVER  BEAMS 

The  best  method  of  deriving  formulas  for  the  deflec- 
tions of  beams  is  by  the  help  of  the  calculus.  These 
methods  are  given  in  higher  works  on  the  subject;  see 


ART.  55  DEFLECTION  OF  CANTILEVER  BEAMS 


107 


for  instance  Merriman's  Mechanics  of  Materials.  The 
formulas  will  be  stated  here  without  proof,  and  be  accom- 
panied by  illustrations  showing  their  value  and  importance. 

When  a  load  P  is  at  the  end  of  a  cantilever  beam  whose 
length  is  /  (Fig.  38),  a  deflection  of  that  end  results,  which 


will  be  designated  by  /.  This  deflection  will  evidently 
be  the  greater  the  greater  the  load  and  the  longer  the 
length  of  the  beam.  The  formula  for  it  is 

f-I* 

J     3EI 

in  which  E  is  the  modulus  of  elasticity  of  the  material 
(Art.  52)  and  7  is  the  moment  of  inertia  of  the  cross- 
section  (Art.  24).  The  ordinate  y  at  any  distance  x  from 
the  free  end  is  given  by  y=lAf(3n  —  n2)  in  which  n  repre- 
sents x/l. 


When  a  uniform  load  is  on  the  beam  let  this  be  called  W 
(Fig.  39).    Then  the  deflection  is 
Wl3 
J     8EI 

It  is  thus  seen  that  the  deflection  varies  as  the  cube  of 
the  length  of  the  beam,  so  that  if  the  length  is  doubled 
the  deflection  will  be  eight  times  as  great.  It  is  also 


108  ELASTIC   DEFORMATIONS  CH.  7 

seen  that  a  uniform  load  produces  only  three-eighths  as 
much  deflection  as  a  single  load  at  the  end.  The  ordinate 
y  at  any  distance  x  from  the  free  end  is  given  by 
y~Vsf  (4n— n4)  where  n  represents  x/l. 

For  example,  let  it  be  required  to  compute  the  deflec- 
tion of  a  cast-iron  cantilever  2X2  inches  and  6  feet  long, 
due  to  a  load  of  100  pounds  at  the  end.  Here  P=100 
pounds,  1  =  72  inches,  ^=15000000  pounds  per  square 
inch,  and  7  =  i/i224  =  li<5  inches4.  Then  from  the  formula, 
/=  0.622  inches,  which  is  the  deflection  at  the  end. 

For  a  rectangular  cross-section  of  breadth  6  and  depth  d 
the  value  of  /  is  y^bd?.  Thus  the  deflections  of  rectangular 
beams  vary  inversely  as  6  and  d3.  As  stiffness  is  the 
reverse  of  deflection,  it  is  seen  that  the  stiffness  of  a  beam 
is  directly  as  its  breadth,  directly  as  the  cube  of  its 
depth,  and  inversely  as  the  cube  of  its  length.  The 
laws  of  stiffness  are  hence  quite  different  from  those  of 
strength. 

Prob.  55  A.  A  steel  I  beam  8  inches  deep  and  6  feet  long  is  used 
as  a  cantilever  to  carry  a  uniform  load  of  240  000  pounds.  What 
will  be  its  deflection? 

Prob.  55  B.  A  cantilever  beam  6.3  feet  long  and  loaded  at  the 
end  has  a  deflection  of  0.48  inches  at  that  end.  What  is  the  deflec- 
tion of  a  point  half-way  between  that  end  and  the  wall? 

ART.  56.     DEFLECTION  OF  SIMPLE  BEAMS 

When  a  simple  beam  of  span  I  has  a  load  P  at  the 
middle  (Fig.  40),  each  reaction  is  %P.  If  this  beam  is 
imagined  to  be  inverted  it  will  be  seen  to  be  equivalent 
to  two  cantilevers  of  length  l/%l,  each  having  the  load  3^P 
at  the  end.  Hence  in  the  first  formula  for  the  deflection 


ART.  56  DEFLECTION  OF  SIMPLE  BEAMS  109 

of  a  cantilever,  given  in  Art.  55,  if  I  is  replaced  by  %l 
and  P  by  %P,  it  becomes 

/_J2L 

J     48EI 

which  gives  the  deflection  of  a  simple  beam  due  to  a  load 
at  the  middle. 


Fig.  40 


When  a  simple  beam  is  loaded  with  w  per  linear  unit 
the  total  load  wl  is  represented  by  W.  The  deflection  at 
the  middle  due  to  this  load  is 


which  is  only  five-eighths  of  the  deflection  caused  by 
the  same  load  at  the  middle. 

The  formulas  of  this  and  the  preceding  article  are 
only  valid  when  the  greatest  horizontal  stress  S  produced 
by  the  load  is  less  than  the  elastic  limit.  These  formulas 
can  be  expressed  in  terms  of  S  by  substituting  the  values 
of  P  and  W  from  the  formula  (4)  of  Art.  28.  Thus  for 
the  simple  beam  with  load  at  the  middle  %Pl=SI/c,  and 
for  the  uniform  load  }/$\7l  =  SI/c.  Hence 

SP 
For  the  single  load  P,         f 


For  the  uniform  load  W,    f=      „ 
4o./iC 

which  show  that  the  deflections  of  beams  under  the 
same  unit-stresses  increase  directly  as  the  squares  of  their 
lengths. 


110  ELASTIC   DEFORMATIONS  CH.  7 

Prob.  56  A.  In  order  to  find  the  modulus  of  elasticity  of  oak,  a 
bar  2X2  inches,  and  6  feet  long,  was  loaded  at  the  middle  with  50 
pounds,  and  then  writh  100  pounds,  the  corresponding  deflections 
being  found  to  be  0.16  and  0.31  inches.  Compute  the  modulus  of 
elasticity  E. 

Prob.  56  B.  Compute  the  deflection  of  a  steel  I  beam  6  inches 
deep  and  16  feet  long  when  it  is  loaded  so  that  the  flexural  unit- 
stress  at  the  middle  equals  the  elastic  limit  of  the  material. 

ART.  57.     RESTRAINED  BEAMS 

A  beam  is  said  to  be  restrained  at  one  end  when  that 
end  is  horizontally  fixed  in  a  wall  and  the  other  end  rests 
on  a  support  (Fig.  41).  In  this  case  the  reaction  of  the 
support  is  less  than  for  a  simple  beam.  For  a  uniform  load 


Fig.  41 

of  w  per  linear  unit  over  the  span  I  it  is  proved  in  Mechan- 
ics of  Materials  that  the  reaction  at  the  support  is 
%wl,  provided  the  elastic  limit  is  not  exceeded.  The 
bending  moment  at  any  section  distant  x  from  the  support 
then  is  %wlx— ^wx2,  and  this  shows  that  when  x  =  %l 
there  is  no  bending  moment;  when  x  =  %l  the  greatest 
positive  bending  moment  is  %2swZ2,  and  when  x  =  l  the 
greatest  negative  bending  moment  is  ^/gwl2;  the  distribu- 
tion of  bending  moments  being  as  shown  in  the  figure. 
Also,  the  maximum  deflection  is 

wl*       JfT^ 
'     185#7     185EI 
which  occurs  when  x  has  the  value  0.42 151. 


ART.  57  RESTRAINED  BEAMS  111 

For  a  beam  fixed  at  both  ends  and  uniformly  loaded 
(Fig.  42)  there  is  a  negative  bending  moment  Vi2w;Z2  at 

-*-->[ 


Fig.  42 

each  wall  and  a  positive  bending  moment  *4ticP  at  the 
middle;  also  the  deflection  at  the  middle  is 

J 


in  which  W  is  the  total  uniform  load  wl. 

It  is  seen  that  in  these  restrained  beams  the  lower  side 
is  partly  in  tension  and  partly  in  compression,  since  a 
positive  bending  moment  indicates  the  former  and  a 
negative  one  the  latter  (Art.  30).  For  a  simple  beam  the 
greatest  bending  moment  is  ]/$Vl,  for  a  beam  fixed  at  both 
ends  the  greatest  bending  moment  is  ~y\%Wl;  hence  if 
both  be  the  same  size  the  restrained  beam  will  carry  the 
greater  load,  or  if  both  carry  the  same  load  the  restrained 
beam  may  be  of  smaller  size  than  the  simple  one.  Thus 
if  beams  can  be  fixed  horizontally  at  their  ends  the  con- 
struction may  be  more  economical.^ 

Prob.  57  A  .  When  a  beam  is  fixed  at  one  end  and  supported  at 
the  other,  the  reaction  of  the  supported  end  due  to  a  load  P  at  the 
middle  is  rAoP-  Show  that  there  is  a  positive  bending  moment 
'fazPl  under  the  load,  and  a  negative  bending  moment  3AePl  at  the 
wall.  Also  draw  the  diagram  of  bending  moments. 

Prob.  57  B.  Show  that  the  deflection  of  a  simple  beam  is  five 
times  as  great  as  that  of  a  beam  fixed  at  both  ends,  both  beams  being 
uniformly  loaded. 


112  ELASTIC  DEFORMATIONS  CH.  7 

ART.  58.    TWIST  IN  SHAFTS 

When  a  shaft  of  length  I  transmits  H  horse-powers  at 
a  speed  of  n  revolutions  per  minute,  one  end  of  the  shaft 
is  twisted  with  respect  to  the  other  through  an  angle  of 
D  degrees,  this  being  the  angle  bed  in  Fig.  34.  When  the 
elastic  limit  is  not  exceeded,  this  angle  is,  for  a  round  shaft, 

TTJ 

D  =  3610000^- 
nFJ 

in  which  J  is  the  polar  moment  of  inertia  of  the  section 
(Art.  46)  and  F  is  the  modulus  of  elasticity  for  shear 
(Art.  52).  Here  I  must  be  taken  in  inches,  J  in  inches4, 
and  F  in  pounds  per  square  inch. 

For  example,  let  a  steel  shaft  125  feet  long,  17  inches 
outside  diameter,  and  11  inches  inside  diameter  transmit 
16  000  horse-powers  at  50  revolutions  per  minute.  Here 
#=16000  horse-powers,  Z=1500  inches,  n  =  50,  F  = 
10000000  pounds  per  square  inch,  ,7  =  6765  inches4. 
Then  from  the  formula  D  =  25.3  degrees,  which  is  the  angle 
through  which  a  point  on  one  end  is  twisted  relative  to  the 
corresponding  point  on  the  other  end.  If  this  shaft 
revolves  with  a  speed  of  only  25  revolutions  per  minute 
while  doing  the  same  work,  its  angle  of  twist  will  be  twice 
as  great  and  the  stresses  in  it  also  twice  as  great  as  before. 
The  formula  also  shows  that  the  angle  of  twist  varies 
directly  as  the  length  of  the  shaft. 

The  above  formula  may  also  be  written 


from   which   D  may  be   computed   when  the   twisting 
moment  Pp  is  given, 


ART.  59  REVIEW  PROBLEMS  113 

Prob.  58  A.  A  solid  steel  shaft  125  feet  long  and  16  inches  in 
diameter  transmits  8000  horse-powers  at  a  speed  of  25  revolutions 
per  minute.  Compute  the  angle  of  twist. 

Prob.  58  B.  A  solid  steel  shaft  8  feet  long  and  2  inches  in  diam- 
eter is  subject  to  the  twisting  moment  brought  by  a  bejt  on  a  pulley 
of  30  inches  diameter,  the  effective  pull  of  the  belt  being  200  pounds. 
Compute  the  angle  of  twist. 

ART.  59.     REVIEW  PROBLEMS 

Prob.  59  A.  Show,  for  timber  and  wrought-iron  bars  stressed 
to  their  elastic  limits,  that  the  change  of  length  cf  the  former  is 
double  that  of  the  latter. 

Prob.  59  B.  Compute  the  tensile  force  required  to  stretch  a  bar 
of  structural  steel,  1%X9%  inches  in  section  area  and  23  feet 
3M  inches  long,  so  that  its  length  may  become  23  feet  3%  inches. 

Prob.  59  C.  Show  that  the  modulus  of  elasticity  is  the  unit-stress 
which  would  stretch  a  bar  to  double  its  original  length,  provided  this 
could  be  done  without  impairing  the  elasticity  of  the  material. 

Prob.  59  D.  What  unit-stress  will  shorten  a  block  of  cast  iron 
0.4  percent  of  its  length? 

Prob.  59  E.  A  cast-iron  bar,  2  inches  wide,  4  inches  deep  and  6 
feet  long,  was  balanced  upon  a  support  and  a  weight  of  4000  pounds 
hung  at  each  end,  when  the  deflection  of  each  end  was  found  to  be 
0.401  inches.  Compute  the  modulus  of  elasticity. 

Prob.  59  F.  Compute  the  elastic  deflection  of  a  light  steel 
10-inch  beam  of  30  feet  span,  due  to  its  own  weight,  when  resting  on 
supports  at  the  ends. 

Prob.  59  G.  Compute  for  the  beam  of  the  last  problem  the 
deflection  when  the  beam  is  fixed  at  both  ends. 

Prob.  59  H.  A  wrought-iron  shaft,  5  feet  long  and  2  inches  in 
diameter,  is  twisted  through  an  angle  of  7  degrees  when  transmitting 
4  horse-powers  at  120  revolutions  per  minute.  Compute  the  shearing 
modulus  of  elasticity. 


114  MISCELLANEOUS  APPLICATIONS  CH.  8 


CHAPTER  8 
MISCELLANEOUS  APPLICATIONS 

ART.  60.     WATER  AND  STEAM  PIPES 

The  pressure  of  water  or  steam  in  a  pipe  is  exerted 
in  every  direction  and  tends  to  tear  the  pipe  apart  longi- 
tudinally. This  external  force  is  resisted  by  the  internal 
tensile  stresses  which  act  in  the  walls  of  the  pipe  normal 
to  the  radii.  If  p  is  the  pressure  per  square  inch  exerted 
by  the  water  or  steam,  D  the  diameter,  and  I  the  length 
of  the  pipe,  the  total  pressure  P  exerted  on  any  diametral 
plane  is  p  .ID.  If  t  is  the  thickness  of  the  pipe  and  S 
the  tensile  unit-stress,  the  total  resisting  stress  will  -be 
S  .  2lt,  if  the  thickness  is  not  large  compared  with  the 
diameter.  Hence 

plD  =  2Slt    or    pD  =  2St 
is  the  formula  for  discussing  water  or  steam  pipes. 

Water  pipes  are  made  of  cast  iron  or  wrought  iron, 
the  former  being  more  common,  while  for  steam  the 
latter  is  preferable.  A  water  pipe  liable  to  the  shock 
of  water-ram  should  have  a  high  factor  of  safety,  and 
in  steam  pipes  the  factors  should  also  be  high.  The 
formula  above  deduced  shows  that  the  thickness  of  a 
pipe  must  increase  with  its  diameter,  as  also  with  the 
internal  pressure  to  which  it  is  to  be  subjected. 

For  example,  let  it  be  required  to  find  the  proper 
thickness  for  a  wrought-iron  steam  pipe  of  18  inches 


ART.  61  RIVETED  LAP  JOINTS  115 

diameter  to  resist  a  steam-pressure  of  250  pounds  per 
square  inch.  With  a  factor  of  safety  of  10  the  working 
unit-stress  is  5500  pounds  per  square  inch.  Then,  from 
the  formula, 

pD     250X18 


so  that  a  thickness  of  ^  inch  would  probably  be  selected. 

The  transverse  resistance  of  a  pipe  is  double  the  longi- 
tudinal, or  pD  =  4St  applies  to  the  case  of  transverse 
resistance.  This  equation  also  applies  to  hollow  spheres 
under  internal  pressure. 

Prob.  60  A.  Find  the  factor  of  safety  of  a  cast-iron  water  pipe 
12  inches  in  diameter  and  Y%  inches  thick  under  a  pressure  of  130 
pounds  per  square  inch. 

Prob.  60  B.  What  internal  pressure  per  square  inch  will  burst  a 
cast-iron  water  pipe  of  24  inches  diameter  and  M  inches  thickness? 

ART.  61.     RIVETED  LAP  JOINTS 

When  two  plates  are  joined  together  by  rivets  and  the 
plates  then  subjected  to  tension,  there  is  brought  a  shear 
upon  the  rivets  which  tends  to  cut  them  off.  A  riveted 
lap  joint  is  one  where  one  plate  simply  laps  over  the  other. 

For  a  lap  joint  with  a  single  row  of  rivets  (Fig.  43), 
let  P  be  the  tensile  stress  which  is  transmitted  from  one 
plate  to  another  by  means  of  one  rivet;  let  a  be  the  pitch 
of  the  rivets,  or  the  distance  from  the  center  of  one  rivet 
to  the  center  of  the  next;  let  d  be  the  diameter  of  a  rivet 
and  t  the  thickness  of  the  plate.  The  plate  tends  to  tear 
apart  on  the  section  area  (a  —  d)t,  while  the  rivet  tends 
to  shear  off  on  the  section  area  %ird?.  Accordingly,  if  St 


116 


MISCELLANEOUS  APPLICATIONS 


CH.  8 


and  S8  are  the  unit-stresses  for  tension  and  shear,  then 

P  =  t(a-d)St      and      P  =  ^7rd2Sa 
are  formulas  for  the  discussion  of  this  case. 

For  example,  a  steel  water  pipe  30  inches  in  diameter 
has  a  longitudinal  rivet  seam  with  one  row  of  rivets, 


Fig.  43 

the  diameter  of  the  rivets  being  %  inches,  their  pitch  2 
inches,  and  the  thickness  of  the  plate  %  inches.  If  the 
interior  water-pressure  is  130  pounds  per  square  inch, 
what  are  the  unit-stresses  in  tension  and  shear?  Here 
the  total  pressure  on  a  diametral  plane  of  length  equal 
to  the  pitch  is  130X2X30  =  7800  pounds.  Then  for 
tension  on  the  plate 

3900 
St  =  T/T^ — 3-77  =  6240  pounds  per  square  inch 

and  for  shear  on  the  rivet 

OQQA 

^« =  n  *7QK\/  JL  =  8860  pounds  per  square  inch 
0.785  X  A 

Here  the  factors  of  safety  are  about  10  for  the  plates 
and  about  6  for  the  rivets,  so  that  the  joint  may  be 
regarded  as  a  satisfactory  one. 


ART.  61 


RIVETED   LAP   JOINTS 


117 


The  thickness  required  for  a  boiler  or  pipe  having  a 
longitudinal  lap  joint  like  that  shown  in  Fig.  43  is  found 
from  the  equation 

2t(a-d)St  =  paD 

where  D  is  the  diameter  and  p  the  inner  unit-pressure. 
For  example,  let  Z)  =  30,  a  =  2,  and  d  =  %  inches,  while 
p=130  pounds  per  square  inch.  Then,  for  ^  =  6000 
pounds  per  square  inch,  the  thickness  is  2=130X2X 
30/2X1^X60000  =  0.52  inches. 

When  two  rows  of  rivets  are  used,  these  are  staggered, 
so  that  the  rivets  in  one  row  come  opposite  the  middle 


Fig.  44 

of  the  pitches  in  the  other  row  (Fig.  44).    Here  the  tension 
P  is  distributed  over  two  rivet  sections  instead  of  one,  and 


P  =  t(a-d)St  F 

are  the  formulas  for  investigation.     Thus  for  the  data 
of  the  above  example,  if  there  are  two  rows  of  rivets, 


118  MISCELLANEOUS  APPLICATIONS  CH.  8 

St  is  12  500  pounds  per  square  inch,  and  $s  =  8900  pounds 
per  square  inch,  which  are  good  working  values  for  mild 
steel  if  the  pipe  is  not  subjected  to  the  shock  of  water-ram. 

The  working  unit-stress  for  shear  should  be  about 
three-fourths  of  that  for  tension,  or  Sa  =  %St.  Equating 
the  above  values  of  P  under  this  condition  gives  a  joint 
where  the  security  of  the  plates  in  tension  is  the  same 
as  that  of  the  rivets  in  shear;  thus 


t 

the  first  being  for  single  lap  riveting  and  the  second  for 
double  lap  riveting.  These  are  approximate  rules  for 
finding  the  pitch  when  the  thickness  of  plates  and  diam- 
eter of  rivets  are  given. 

Prob.  61-4.  A  steel  water  pipe  30  inches  in  diameter  has  rivets 
%  inches  in  diameter  and  plates  J/£  inches  thick.  If  double  riveting 
is  used,  what  should  be  the  pitch  of  the  rivets? 

Prob.  61  B.  Compute  the  factor  of  safety  of  a  steel  boiler  5  feet 
in  diameter  and  -y\Q  inches  thick  when  it  is  subject  to  a  steam 
pressure  of  300  pounds  per  square  inch,  there  being  longitudinal 
lap  joints  having  rivets  of  %  inches  diameter  with  2^  inches  pitch. 

ART.  62.     RIVETED  BUTT  JOINTS 

When  two  plates  butt  together  cover  plates  are  used 
on  one  or  both  sides;  if  the  covers  are  on  both  sides  each 


Fig.  45 

is  one-half  the  thickness  of  the  main  plate  (Fig.  45). 
The  shear  on  each  rivet  is  here  divided  between  the  upper 
and  the  lower  cross-section,  this  being  called  a  case  of 


ART.  62  RIVETED  BUTT  JOINTS  119 

double  shear.  Thus  if  P  is  the  tension  which  is  trans- 
mitted through  one  rivet,  d  the  diameter  of  a  rivet,  a  the 
pitch,  and  t  the  thickness  of  the  main  plate, 

P  =  t(a- d)St  P  =  2.  }4ird2S8 

which  are  the  same  as  for  two  rows  of  lap  riveting. 

The  'efficiency'  of  a  riveted  joint  is  the  ratio  of  the 
strength  of  the  joint  to  that  of  the  solid  plate.  If  the 
joint  is  designed  so  as  to  be  of  equal  strength  in  tension 
and  shear  this  efficiency  is 

t(a-d)St=       d 
taSt  a 

Thus  if  the  rivets  are  %  inches  in  diameter  and  the  pitch 
is  2  inches  the  efficiency  is  1-%  =  0.625,  that  is,  the 
riveted  joint  has  only  62.5  percent  of  the  strength  of  the 
solid  plate.  Single  lap  riveting  has  usually  an  efficiency 
of  about  60  percent,  while  double  lap  riveting  and  common 
butt  riveting  have  from  70  to  75  percent.  By  using  three 
or  more  rows  of  rivets  efficiencies  of  over  80  percent 
can  be  secured. 

When  a  joint  is  not  of  equal  strength  in  tension  and 
shear  there  are  two  efficiencies,  one  being  the  ratio  of  the 
tensile  strength  of  the  joint  to  that  of  the  solid  plate, 
and  the  other  the  ratio  of  the  shearing  strength  to  that  of 
the  solid  plate.  The  least  of  these  is  the  true  efficiency. 

Prob.  62^1.  A  butt  joint  with  two  cover-plates  has  the  main 
plate  Yi  inches  thick,  the  rivets  %  inches  in  diameter,  and  the  pitch 
of  the  rivets  2%  inches.  Compute  the  efficiency. 

Prob.  62  B.  Show  that  the  efficiency  of  a  butt  joint,  based  on 
the  shear  of  rivets,  is  double  that  of  a  lap  joint. 


120  MISCELLANEOUS   APPLICATIONS  CH.  8 

ART.  63.    STRESSES  DUE  TO  TEMPERATURE 

A  bar  which  is  free  to  move  elongates  when  the  tem- 
perature rises  and  shortens  when  it  falls.  But  if  the  bar 
is  under  stress,  or  is  fixed  so  that  it  cannot  elongate  or 
contract,  the  change  in  temperature  produces  a  certain 
unit-stress.  This  unit-stress  is  that  which  would  cause 
a  change  in  length  equal  to  that  produced  in  the  free  bar 
by  the  change  in  temperature. 

The  coefficient  of  linear  expansion  is  the  elongation 
of  a  bar  of  length  unity  under  a  rise  of  temperature  of 
one  degree.    For  the  Fahrenheit  degree  the  average  values 
of  the  coefficients  of  expansion  are  as  follows: 
For  brick  and  stone,    (7  =  0.0000050 
For  cast  iron,  C  =  0.0000062 

For  wrought  iron,        C  =  0.0000067 
For  steel,  C  =  0.0000065 

Thus  a  free  bar  of  cast  iron  1000  inches  long  will  elongate 
0.0062  inches  for  a  rise  of  one  degree,  and  0.62  inches 
for  a  rise  of  100  degrees. 

The  elongation  of  a  bar  of  length  unity  for  a  change 
of  t  degrees  is  hence  s  =  Ct.  But  (Art.  52)  the  unit-stress 
due  to  the  unit-elongation  s  is  S  =  sE,  where  E  is  the 
modulus  of  elasticity.  Therefore 

S  =  CtE 

is  the  unit-stress  produced  by  a  change  of  t  degrees  on 
a  bar  which  is  fixed.  If  the  temperature  rises  S  is  com- 
pression, if  the  temperature  falls  S  is  tension. 

For  example,  consider  a  wrought-iron  rod  which  is 
used  to  tie  together  two  walls  of  a  building  and  which  is 


ART.  64  SHRINKAGE  OF  HOOPS  121 

screwed  up  to  a  stress  of  10  000  pounds  per  square  inch. 
If  the  temperature  falls  50  degrees  there  is  produced  a 
tensile  unit-stress, 

S  =  0.0000067  X  50  X  25  000  000  =  8400 
and  hence  the  total  stress  in  the  rod  is  18  400  pounds 
per  square  inch.  If  the  temperature  rises  50  degrees 
the  stress  in  the  bar  is  reduced  to  1600  pounds  per  square 
inch.  In  all  cases  the  unit-stresses  due  to  temperature 
are  independent  of  the  length  and  section  area  of  the  bar. 

Prob.  63  A.  A  cast-iron  bar  6  feet  long  and  4X4  inches  in  section 
is  confined  between  two  immovable  walls.  What  pressure  is  brought 
on  the  walls  by  a  rise  of  40  degrees  in  temperature? 

Prob.  63  B.  When  steel  railroad  rails  are  improperly  laid  with 
their  ends  close  together  at  a  temperature  of  40°,  what  compressive 
unit-stress  occurs  when  the  temperature  rises  to  80°? 

ART.  64.  SHRINKAGE  OF  HOOPS 
A  hoop  or  tire  is  frequently  turned  with  the  inner 
diameter  slightly  less  than  that  of  the  cylinder  or  wheel 
upon  which  it  is  to  be  placed.  The  hoop  is  then  expanded 
by  heat  and  placed  upon  the  cylinder,  and  upon  cooling 
it  is  held  firmly  in  position  by  the  radial  stress  produced. 
This  radial  stress  causes  tension  in  the  hoop. 

Let  D  be  the  diameter  of  the  cylinder  upon  which 
the  hoop  is  to  be  shrunk  and  d  the  interior  diameter  to 
which  the  hoop  is  turned.  If  the  thickness  of  the  hoop 
is  small,  D  will  be  unchanged  by  the  shrinkage  and  d 
will  be  increased  to  Z).  The  unit-elongation  of  the  hoop 
is  then  s=  (D  —  d)/d,  and  hence  the  unit-stress  produced  is 


where  E  is  the  modulus  of  elasticity  of  the  material. 


122  MISCELLANEOUS  APPLICATIONS  CH.  8 

A  common  rule  in  steel-hoop  shrinkage  is  to  make 
D—d  equal  to  Visood;  that  is,  the  cylinder  is  turned  so 
that  its  diameter  is  Visooth  greater  than  the  inner  diameter 
of  the  hoop.  Accordingly  the  tangential  unit-stress  which 
occurs  in  the  hoop  after  shrinkage  is  30000000/1500  = 
20  000  pounds  per  square  inch. 

When  the  hoop  is  thick  the  above  rule  is  not  correct, 
for  a  part  of  the  stress  produced  by  the  shrinkage  causes 
the  diameter  of  the  cylinder  to  be  decreased.  The  rules 
for  this  case  are  complex  ones,  and  cannot  be  developed 
in  an  elementary  text-book;  they  will  be  found  in  Chapter 
XVI  of  Mechanics  of  Materials. 

Prob.  64.  Upon  a  cylinder  18  inches  in  diameter  a  thin  wrought- 
iron  hoop  is  to  be  placed.  The  hoop  is  turned  to  an  inner  diameter 
of  17.98  inches  and  then  shrunk  on.  Compute  the  tensile  unit- 
stress  in  the  hoop. 

ART.  65.     SHAFT  COUPLINGS 

Let  a  shaft  be  in  two  parts  which  are  connected  by 
a  flange  coupling.  (Fig.  46).  A  shows  the  end  view 


and  B  the  side  view  of  the  coupling.  The  flanges  of  the 
coupling  are  connected  by  bolts  which  are  brought  into 
shearing  stress  in  transmitting  the  torsion  from  one  part 
of  the  shaft  to  the  other. 

Let  the  shaft  be  solid  and  of  diameter  Z),  let  there 
be  n  bolts  of  diameter  d,  and  let  h  be  the  distance  from 


ART.  66        RUPTURE  OF  BEAMS  AND  SHAFTS  123 

the  center  of  the  shaft  to  the  center  of  a  bolt.  If  D  and 
d  are  assumed,  as  also  the  distance  h,  then,  as  shown  in 
Art.  124  of  Mechanics  of  Materials,  the  formula 

_  (d+2h)Ds 


gives  the  number  of  bolts  which  is  required  in  order  that 
the  strength  of  the  bolts  may  be  the  same  as  the  strength 
of  the  shaft.  Since  d  is  usually  much  smaller  than  h 
it  may  be  neglected  within  the  parentheses;  then  the 
above  formula  becomes  n  =  D3/4ihd2  which  is  a  simpler 
expression  generally  used  in  practice. 

For  example,  let  D  =  8  inches,  d=l  inch,  and  h=l2 
inches,  then  the  second  formula  gives  n=10.7,  so  that 
11  bolts  should  be  used.  If  D  =  8  inches,  d=l%  inches, 
and  h  =  12  inches,  the  formula  gives  n  =  6.8  so  that  seven 
bolts  should  be  used. 

The  case  shown  at  CD  in  the  above  figure  is  one  that 
should  never  occur  in  practice,  because  here  the  bolts 
are  subject  to  a  bending  stress  as  well  as  to  the  shearing 
stresses  due  to  the  torsion.  It  is  clear  that  this  bending 
stress  will  increase  with  the  length  between  the  flanges, 
and  that  the  bolts  should  be  greater  in  diameter  than 
for  the  case  of  pure  shearing. 

Prob.  65.  A  solid  steel  shaft  16  inches  in  diameter  transmits 
16  000  horse-powers  at  50  revolutions  per  minute.  Design  a  flange 
coupling  for  this  shaft. 

ART.  06.  RUPTURE  OF  BEAMS  AND  SHAFTS 

The  formulas  (4)  and  (6)  deduced  in  Arts.  28  and  47 
for  the  discussion  of  beams  and  shafts  are  only  valid 
when  the  unit-stress  S  is  less  than  the  elastic  limit  of 


124 


MISCELLANEOUS   APPLICATIONS 


CH.  8 


the  material.  Formula  (4)  can,  however,  be  used  for 
the  rupture  of  beams,  if  S  be  taken  as  a  certain  quan- 
tity intermediate  between  the  ultimate  compressive  and 
tensile  strengths  of  the  material.  This  quantity,  which  is 
called  the  'modulus  of  rupture/  has  been  determined 
by  breaking  beams  under  transverse  loads  and  then 
computing  S  from  the  formula.  In  like  manner  formula 
(6)  can  be  used  for  the  rupture  of  round  shafts  if  the 
modulus  of  rupture,  as  found  by  experiment,  is  used 
instead  of  the  ultimate  shearing  strength. 

The  following  table  gives  average  values  of  the  modulus 
of  rupture  as  determined  by  testing  beams  and  columns 

TABLE  11.     MODULUSES  OF  RUPTURE 


Material 

For  Beams 

For  Round  Shafts 

Pounds 
per  Square 
Inch 

Kilos 
per  Square 
Centimeter 

Pounds 
per  Square 
Inch 

Kilos 
per  Square 
Centimeter 

Timber 

9000 

630 

2000 

140 

Stone 

2000 

140 

Cast  Iron 

35000 

2450 

25  000 

1750 

Wrought  Iron 
Steel,  hard 

80000 

5600 

40000 
70000 

2800 
4900 

to  destruction.  Wrought  iron  and  mild  steel  have  no 
proper  modulus  of  rupture  when  used  as  beams,  since 
they  continually  bend  and  do  not  break  until  the  ulti- 
mate load  is  reached.  By  the  use  of  these  values  formulas 
(4)  and  (6)  may  be  applied  to  the  solution  of  numerous 
problems  relating  to  the  rupture  of  beams  and  shafts. 

For  example,  let  it  be  required  to  find  the  size  of  a 
square  cast-iron  simple  beam  of  6  feet  span  that  will 
rupture  under  its  own  weight.  Let  x  be  the  side  of  the 


ART.  67  REVIEW  PROBLEMS  125 

square.  The  weight  of  a  cast-iron  bar  x  square  inches 
in  section  area  and  one  yard  long  is  9 Ax  pounds;  thus  the 
weight  of  the  beam  is  I8.8x  pounds.  The  bending  moment 
is  %Wl  or  169.2z  pound-inches.  The  value  of  c  is  Y^x 
and  that  of  I  is  1/i2^4-  Then,  by  formula  (4), 
35000X*' 

6x 
and  the  solution  of  this  gives  a:  =  0.1 7  inches. 

It  is  to  be  noted,  when  formulas  (4)  and  (6)  are  used 
for  cases  of  rupture,  that  they  are  entirely  empirical 
and  have  no  rational  basis. 

Prob.  66  A.  What  load  P  applied  at  the  middle  of  a  cast-iron 
bar,  18  inches  long  and  1  inch  square,  will  cause  its  rupture? 

Prob.  66  B.  What  force  P,  acting  at  24  inches  from  the  axis 
of  a  steel  shaft  1.4  inches  in  diameter,  will  cause  failure  by  torsion? 

ART.  67.     REVIEW  PROBLEMS 

Prob.  67  A.  What  internal  pressure  per  square  inch  will  burst  a 
cast-iron  sphere  24  inches  in  diameter  and  %  inches  thick? 

Prob.  67  B.  A  wrought-iron  boiler,  63  inches  in  diameter  and 
1:Vio  inches  thick,  carries  a  steam  pressure  of  180  pounds  per  square 
inch.  Find  the  factor  of  safety  of  the  metal  when  the  efficiency 
of  the  longitudinal  riveted  joint  is  87  percent. 

Prob.  67  C.  Draw  a  figure  for  a  double-riveted  butt  joint  and 
deduce  formulas  for  the  same.  Find  the  efficiencies  for  plate  and 
rivets  when  plate  thickness  is  Y^  inch,  pitch  of  rivets  is  334  inches, 
and  diameter  of  rivets  is  1%e  inches. 

Prob.  67  D.  Find  the  radial  unit-pressure  between  the  rim  and 
tire  of  a  locomotive  driving-wheel  when  the  shrinkage  is  Visoo,  the 
diameter  of  the  tire  being  60  inches  and  its  thickness  %  inches. 

Prob.  67  E.  A  solid  shaft  6  inches  in  diameter  is  coupled  by  bolts 
13/t  inches  in  diameter  with  their  centers  5  inches  from  the  axis. 
How  many  bolts  are  necessary? 


126  REINFORCED    CONCRETE  CH.  9 

CHAPTER  9 

REINFORCED   CONCRETE 
ART.  68.     CONCRETE  AND  STEEL 

Columns  and  beams  are  made  by  ramming  concrete 
into  wooden  forms  or  boxes  which  surround  their  sides, 
these  being  removed  after  the  concrete  has  hardened. 
Steel  rods  are  often  placed  in  the  forms  and  the  concrete 
rammed  around  them,  and  this  combination  is  often 
called  'reinforced  concrete.'  The  object  of  inserting  the 
steel  is  to  make  a  safer  and  stronger  construction  than  is 
possible  with  concrete  alone,  and  to  do  this  at  a  lower 
cost  than  is  possible  when  only  steel  is  used.  Since  1895 
there  has  been  a  great  development  in  this  kind  of  con- 
struction, steel  rods  being  now  extensively  used  for  the 
reinforcement  not  only  of  columns  and  beams  but  also 
in  walls,  sewers,  and  arches. 

The  kind  of  concrete  generally  used  for  this  purpose 
is  made  of  Portland  cement,  sand,  and  broken  stone, 
an  excellent  grade  being  of  the  proportions  1  cement, 
2  sand,  4  stone  by  measure,  and  a  lower  grade  being 
of  the  proportions  1  cement,  3  sand,  6  stone;  these  two 
grades  are  frequently  called  '1  :  2  :  4  concrete '  and 
'1  :  3  :  6  concrete ' ;  the  former  is  stronger  and  better  than 
the  latter,  but  its  cost  is  higher.  The  average  weight  of 
concrete  is  about  150  pounds  per  cubic  foot,  or  about 
the  same  as  that  of  sandstone. 

The  strength  of  concrete  increases  with  its  age,  reaching 


ART.  68  CONCRETE  AND  STEEL  127 

nearly  the  highest  value  by  the  end  of  the  first  year.  Its 
ultimate  compressive  strength  is  much  higher  than  the 
ultimate  tensile  strength,  and  the  following  are  average 
values  of  these  in  pounds  per  square  inch  for  concrete 
one  year  old: 

Compressive  Tensile 

Strength  Strength 

For  1  : 2  : 4  concrete  3500  300 

For  1  :  3  :  6  concrete  2500  200 

These  figures,  like  all  those  given  in  this  chapter,  refer 
only  to  concrete  made  with  Portland  cement.  The  ulti- 
mate shearing  strength  of  concrete  is  from  800  to  1000 
pounds  per  square  inch. 

It  is  seen  from  these  values  of  the  ultimate  strengths 
that  concrete  is  not  well  adapted  to  resist  tension,  and 
under  tensile  stresses  it  is  almost  impracticable  to  use  con- 
crete, unless  it  be  strengthened  by  reinforcing  rods  of 
metal. 

Concrete  suffers  a  greater  change  of  shape  under  a 
given  applied  unit-stress  than  steel,  or  the  stiffness  of 
steel  is  much  greater  than  that  of  concrete.  Mean  values 
of  the  modulus  of  elasticity  for  the  two  grades  of  concrete 
are  in  pounds  per  square  inch: 

For  1  :  2  :  4  concrete  E  =  3  000  000 

For  1  :  3  :  6  concrete  E  =  2  000  000 

For  steel  the  mean  value  of  E  is  30  000  000  pounds  per 
square  inch  (Art.  52),  and  hence  it  is  seen  that  concrete 
suffers  an  elastic  deformation  ten  or  fifteen  times  as  great 
as  that  of  steel  when  subjected  to  the  same  stress  per 
square  inch. 


128  REINFORCED    CONCRETE  CH.  9 

The  elastic  limit  of  concrete  is  not  well  defined,  but  as 
a  rough  average  it  may  be  taken  in  compression  at  about 
one-sixth  and  in  tension  at  about  one-fifth  of  the  ultimate 
strength.  The  allowable  working  unit-stress  for  concrete 
under  compression  is  generally  taken  as  about  one- 
seventh  of  the  ultimate  strength,  that  is,  about  500 
pounds  per  square  inch  for  1:2:4  concrete  and  about 
350  pounds  per  square  inch  for  1:3:6  concrete. 

The  rods  or  bars  used  for  reinforcement  are  generally 
of  structural  steel  having  an  ultimate  strength  of  about 
60  000  and  an  elastic  limit  of  about  35  000  pounds  per 
square  inch.  These  may  be  the  round,  square,  and 
rectangular  shapes  such  as  are  everywhere  found  in  the 
market,  and  various  special  patented  forms  are  also  widely 
used.  The  Ransome  rods  are  square  bars  which  have  been 
twisted  so  that  the  corner  lines  are  spirals;  the  Thacher 
and  Johnson  bars  are  rolled  so  as  to  have  protuberances 
or  swellings  at  intervals  along  the  length;  these  forms  are 
claimed  to  possess  special  advantages  in  preventing  the 
rods  from  slipping  in  the  concrete.  The  Kahn  bar  has 
projecting  fins  which  are  intended  to  prevent  beams  from 
shearing,  and  there  are  also  several  other  forms  which  are 
advertised  and  used.  Some  of  these  are  claimed  to  be 
advantageous  in  having  an  elastic  limit  much  higher  than 
that  of  structural  steel. 

Prob.  68^1.  Consult  the  advertising  columns  of  the  engineering 
journals  and  obtain  pictures  of  several  kinds  of  reinforcing  bars 
for  beams. 

Prob.  68B.  Compute  the  amount  of  shortening  of  a  1  :  2  :  4 
concrete  column,  12  inches  square  and  9  feet  long,  under  a  load 
of  60  000  pounds. 


ART.  69  COMPOUND  BARS  129 

ART.  69.     COMPOUND  BARS 

A  bar  which  is  under  tension  is  usually  called  a  rod, 
while  one  under  compression  is  called  a  strut.  The  rods 
and  struts  considered  in  the  previous  chapters  may  be 
called  simple  ones,  as  each  has  been  only  of  one  material; 
a  compound  rod  or  strut,  however,  is  one  formed  of  two 
or  more  kinds  of  materials.  For  example,  a  steel  rope 
with  a  hempen  center,  or  a  concrete  column  with  a  steel 
bar  within  it  parallel  to  the  axis,  are  instances  of  com- 
pound tension  and  compression  members. 

When  a  rod  of  two  materials  is  subject  to  a  longitudinal 
tension  P,  part  of  this  is  resisted  by  one  material  and 
part  by  the  other.  Let  A\  be  the  section  area  of  one 
material  and  A*  that  of  the  other,  and  let  Si  be  the  unit- 
stress  over  the  area  AI,  and  Sz  the  unit-stress  over  the 
area  A2.  Then  AiSi  and  A2S2  are  the  total  stresses  on 
the  two  sections,  and  hence,  since  the  resisting  stresses 
must  equal  the  applied  tension, 

P  =  AiSi+AzSz  (9) 

is  a  necessary  equation  of  equilibrium.  When  P,  AI,  and 
AZ  are  given,  the  values  of  Si  and  S2  cannot  be  determined 
from  this  equation,  and  hence  a  second  condition  between 
them  must  be  derived. 

This  second  condition  is  established  from  the  fact 
that  the  elongation  of  the  two  parts  of  the  bar  is  the  same. 
Let  EI  be  the  modulus  of  elasticity  of  the  first  material 
and  Ez  that  of  the  second.  Then,  if  the  elastic  limit 
of  neither  material  is  exceeded,  the  elongation  of  a  unit 
of  length  of  the  first  material  is  Si/Ei  (Art.  52)  and  that 


130  REINFORCED    CONCRETE  CH.  9 

of  the  second  material  is  S2/E2.     Since  these  must  be 
equal, 

S\        $2  Ol        El 


which  shows  that  the  unit-stresses  in  the  two  materials 
are  proportional  to  their  moduluses  of  elasticity.  If 
Ez  is  10  times  as  great  as  E\,  then  S2  must  be  10  times 
as  great  as  Si. 

By  help  of  the  above  formulas  the  values  of  Si  and  Sz 
due  to  a  load  P  on  a  compound  rod  of  two  materials  may 
be  easily  found.  The  above  reasoning  applies  also  to 
compression  if  the  length  of  the  strut  is  not  more  than 
about  10  times  its  thickness,  so  that  lateral  flexure  does 
not  change  the  uniform  distribution  of  the  stresses.  For 
example,  consider  a  wooden  bar  having  wrought-iron 
straps  fastened  along  two  opposite  sides;  here  E\  for  the 
timber  is  1  500  000,  while  Ez  for  the  wrought  iron  is 
25  000  000  pounds  per  square  inch,  so  that  Ei/E2  is  0.06, 
and  hence  the  unit-stress  Si  in  the  timber  is  equal  to 
0.06$2,  so  that,  if  S2  is  5000  pounds  per  square  inch,  Si 
will  be  300  pounds  per  square  inch.  The  formula  Si/S%  = 
Ei/Ez  cannot,  however,  be  used  when  Si  exceeds  the 
elastic  limit  of  the  timber  or  when  $2  exceeds  the  elastic 
limit  of  the  wrought  iron. 

The  determination  of  the  values  of  Si  and  $2  for  this 
compound  rod  is  now  easily  made  when  AI,  At,  and  P 
are  given.  Let  AI  for  the  timber  be  30  square  inches  and 
A2  for  the  wrought  iron  be  4  square  inches.  Let  the  load 
P  be  60  000  pounds.  Since  £i  =  0.06£2,  formula  (9)  gives 

60000  =  36X0.06^2+4^2 
from  which  $2  =  9740  pounds  per  square  inch  for  the 


ART.  70 


REINFORCED    COLUMNS 


131 


wrought  iron,  and  hence  $i  =  580  pounds  per  square  inch 
for  the  timber.  It  is  here  seen  that  the  total  stress  which 
comes  on  the  wrought  iron  is  4X9740  or  about  39  000 
pounds,  while  that  on  the  timber  is  about  21  000  pounds. 
The  metal  hence  carries  the  greater  part  of  the  load, 
and  it  does  this  largely  by  virtue  of  its  greater  stiffness. 
In  general,  the  higher  the  value  of  E  for  a  material  in  a 
compound  bar,  the  greater  is  the  part  of  the  load  carried 
by  it. 

Prob.  69.  A  short  timber  strut,  8X8  inches  in  section,  has  four 
steel  plates  fastened  to  its  sides,  each  being  6  X  /^  inches  in  size,  and 
it  carries  a  load  of  180  000  pounds.  Compute  the  compressive  unit- 
stresses  in  the  two  materials. 

ART.  70.     REINFORCED  COLUMNS 

Concrete  columns  are  generally  square  or  round. 
Fig.  47  shows  a  square  form  having  four  steel  rods  im- 


>»., 


Fig.  49 


Fig.  47  Fig.  48 

bedded  in  it  near  the  corners.  Fig.  48  shows  a  round 
column  having  a  single  rod  through  its  axis.  Fig.  49 
shows  a  round  column  formed  by  a  hollow  metal  cylinder 
which  is  filled  with  concrete.  The  length  of  these  columns 
will  be  considered  short  compared  with  the  thickness, 
so  that  no  tendency  to  lateral  flexure  exists. 


132  REINFORCED    CONCRETE  CH.  9 

The  investigation  of  a  reinforced  column  consists  in 
determining  the  compressive  unit-stresses  due  to  the 
given  load  P,  and  comparing  them  with  the  allowable 
unit-stresses.  Let  A\  be  the  section  area  of  the  concrete 
and  Az  that  of  the  metal,  and  let  E\  and  E2  be  the  corre- 
sponding moduluses  of  elasticity.  Let  n  represent  the 
known  ratio  Ez/E\\  then,  as  shown  in  the  last  article, 
the  value  of  Sz/Si  must  also  be  n,  provided  the  elastic 
limit  of  neither  material  is  exceeded.  Inserting  for  $2 
its  value  nSi  in  the  first  formula  of  Art.  69  and  solving, 
there  results 


from  which  the  unit-stress  in  the  concrete  may  be  com- 
puted, and  then  the  unit-stress  in  the  metal  is  found 
from  Sz  =  nSi. 

As  an  example  of  investigation,  take  a  reinforced 
column  of  1  :  3  :  6  concrete  which  is  14  inches  square  and 
has  four  steel  rods,  each  %  inches  in  diameter,  parallel 
with  its  length  near  the  corners,  as  in  Fig.  47,  while  the 
load  P  is  71  000  pounds.  Here  the  section  area  of  the 
four  rods  is  A2  =  0.442  square  inches  and  that  of  the  con- 
crete is  ^li  =  196  —  0.44=195.56  square  inches.  Then, 
since  n  is  15  (Art.  68),  formula  (9)'  -gives  Si  =  7l  000/202.2 
=  351  pounds  per  square  inch  for  the  concrete,  and  hence 
$2  =  5270  pounds  per  square  inch  for  the  steel.  These  are 
safe  working  stresses,  that  for  the  steel  being  very  low. 

When  a  reinforced  column  is  to  be  designed,  the  load  P 
and  the  unit-stress  Si  for  the  concrete  are  known  or 
assumed,  while  n  is  to  be  taken  as  10  or  15,  depending 
on  the  kind  of  concrete  (Art.  68).  Then  the  above 


ART.  70  REINFORCED  COLUMNS  133 

investigation  shows  that  the  section  areas  AI  and  A2 
of  the  two  materials  are  to  be  determined  so  as  to  satisfy 
the  equation 

p 


In  order  to  do  this,  one  section  area  is  usually  assumed 
and  the  other  can  then  be  computed.  Evidently  many 
different  sets  of  values  of  AI  and  A2  can  be  found,  and  the 
one  to  be  used  will  generally  be  determined  by  conven- 
ience and  local  conditions.  For  example,  let  the  column 
in  Fig.  48  be  of  1:2:4  concrete  for  which  Si  is  to  be 
500  pounds  per  square  inch,  and  it  be  required  that  the 
diameter  shall  be  6  inches,  while  the  load  P  is  20  000 
pounds.  Let  it  be  required  to  find  the  diameter  d  of  the 
single  steel  rod.  Here  A^^ird2,  and  Ai  =  ]4ir(3G-d2), 
and  then 

Y±TT  (36  -  d  -  +  1  Orf2)  =  20  000/500 

from  which  d  is  found  to  be  1.28  inches,  so  that  a  rod 
1%6  inches  in  diameter  should  be  used. 

For  a  column  like  Fig.  49,  cast  iron  is  often  used  for 
the  outer  casing,  and  since  E  is  15  000  000  pounds  per 
square  inch  for  cast  iron,  the  value  to  be  used  for  n  will 
be  about  7j^  for  1:3:6  concrete,  which  is  the  grade  that 
would  be  most  likely  to  be  used  for  a  column  of  this  kind. 
The  safe  load  P  which  such  a  column  can  carry  may  then 
be  found  by  formula  (9),  in  which  Si  =  350  and  S2  = 
73^X350  =  2620  pounds  per  square  inch.  The  inves- 
tigation or  design  of  such  a  column  may  be  made  by  the 
methods  above  explained. 

Prob.  70.  A  column  like  Fig.  49  is  to  be  made  as  described 
in  the  last  paragraph.  The  diameter  of  the  concrete  is  2  feet  6  inches. 


134 


REINFORCED    CONCRETE 


CH.  9 


and  the  load  to  be  carried  is  500  000  pounds.    Compute  the  thick- 
ness of  the  cast-iron  casing. 

ART.  71.    BEAMS  WITH  SYMMETRIC  REINFORCEMENT 

Although  beams  of  timber  are  sometimes  reinforced  by 
fastening  metal  plates  upon  the  sides,  the  most  common 
example  of  reinforcement  is  that  of  a  concrete  beam  in 
which  the  steel  is  imbedded.  Concrete  beams  are 
usually  rectangular  and  Fig.  50  represents  a  section  of 
one  with  no  reinforcing  rods,  this  being  called  a  plain 
concrete  beam;  such  a  beam  is  not  well  adapted  for 
carrying  heavy  loads  on  account  of  the  low  tensile  strength 
of  the  concrete  (Art.  68).  Fig.  51  gives  a  section  of  a 


Fig.  50 


Fig.  51 


Fig.  52 


concrete  beam  in  which  a  steel  I  beam  is  imbedded; 
although  this  form  is  sometimes  used,  the  steel  part  is 
generally  made  sufficiently  large  to  carry  the  given  loads, 
the  office  of  the  concrete  being  to  protect  the  steel  from 
fire.  Fig.  52  is  a  form  in  which  steel  rods  are  imbedded 
near  both  the  top  and  the  bottom  of  the  beam,  and 
symmetrically  arranged  with  respect  to  the  neutral  axis 
of  the  concrete  section. 

The  discussion  of  a  plain  concrete  beam  is  made  by  the 
help  of  formula  (4)  and  the  methods  explained  in  Arts. 
28-31.  For  example,  let  the  beam  be  8  inches  wide,  10 


ART.  71      BEAMS  WITH  SYMMETRIC   REINFORCEMENT      135 

inches  deep,  6  feet  in  span  and  let  it  be  required  to  compute 
the  total  uniform  load  which  it  can  carry  when  the  concrete 
is  stressed  to  100  pounds  per  square  inch  on  the  tensile 
side.  The  bending  moment  for  the  total  load  W  is 
^sWX72  in  inch-pounds  (Art.  21).  The  section  modulus 
I/c  is,  from  Arts.  23  and  24,  found  to  be  y0X8X102  = 
133.3  inches3.  Accordingly,  the  flexure  formula  (4)  gives 

M  =  S(I/c),       or       QW=  100X133.3 

from  which  W=1480  pounds  is  the  total  uniform  load 
which  the  beam  can  carry.  Since  this  beam  weighs  about 
500  pounds,  an  additional  uniform  load  of  about  980 
pounds  will  stress  the  concrete  on  the  tensile  side  to  100 
pounds  per  square  inch. 

For  cases  where  the  imbedded  steel  is  placed  sym- 
metrically with  respect  to  the  concrete  section  as  in  Figs. 
51  and  52,  the  flexure  formula  (4)  may  be  modified  so  as 
to  take  the  two  different  materials  into  account.  Let 
S\  be  the  unit-stress  in  the  concrete  on  the  remotest  fiber 
at  the  distance  Ci  from  the  neutral  axis,  $2  the  unit-stress 
in  the  steel  at  the  distance  £2,  and  /i  and  72  the  moments 
of  inertia  of  the  concrete  and  steel  sections.  M  being 
the  maximum  bending  moment  carried  by  the  beam, 
this  must  be  equal  to  the  sum  of  the  resisting  moments 
of  the  two  sections  (Art.  26).  Therefore,  for  compound 
beams, 


Now,  for  the  beam  as  for  the  column,  the  change  of 
length  of  any  line  drawn  on  the  sides  parallel  with  the 
length  must  be  the  same  for  both  concrete  and  steel. 
At  the  distance  unity  from  the  neutral  axis  the  unit- 


136  REINFORCED    CONCRETE  CH.  9 

stress  in  the  concrete  is  Si/ci  and  the  change  in  a  unit 
of  length  is  then  Si/CiEi,  if  EI  is  the  modulus  of  elasticity 
of  the  concrete.  Similarly  for  the  steel  the  change  in  a 
unit  of  length  at  the  distance  unity  from  the  neutral  axis 
is  Sz/c2E2.  Hence 

S        S  c 

—  TT  =  -~  ,       whence       S2  =  -nS< 
c1El     c2E2>  '     c, 

in  which  n  represents  the  ratio  E2/Ei,  the  value  of  which 
is  10  or  15  (Art.  68).  Also  inserting  this  value  of  $2 
in  the  above  expression  for  M,  there  results 

or      ft-  dO) 


which  are  formulas  for  the  design  and  investigation  of 
reinforced  concrete  beams. 

As  an  example,  let  it  be  required  to  find  the  total 
uniform  load  W  which  a  beam  like  Fig.  52  can  carry, 
Si  being  100  pounds  per  square  inch  and  n  being  10. 
Let  the  width  be  8  inches,  the  depth  10  inches,  the  span 
6  feet,  and  each  of  the  six  steel  rods  be  %  inch  in  diameter 
and  have  its  center  3  inches  ifrom  the  neutral  axis.  The 
value  of  M  in  pound-inches  is  ]/^WX12,  or  9W.  The 
section  area  of  each  rod  is  0.196  square  inches,  so  that  the 
moment  of  inertia  72  is  6X0.196X32=  10.6  inches4.  The 
moment  of  inertia  of  the  8  X  10-inch  rectangle  is  1/12X8 
X103  =  666.7  inches4,  and  hence  h  =  666.7  -10.6  =  656 
inches4.  Then,  since  Si/c\  is  100/5,  the  first  formula  in 
(10)  gives  QW  =  20(656+  106),  from  which  W=172Q 
pounds,  which  is  13  percent  more  than  a  plain  concrete 
beam  of  the  same  dimensions  can  carry. 

For  the  above  case  the  unit-stress  82  in  the  steel  rod 
is  quite  small.  If  the  rods  are  at  the  upper  and  lower 


ART.  72         UNSYMMETRIC  REINFORCEMENT  137 

surfaces  of  the  beam,  the  above  formula  shows  that  S2 
is  nSi,  but  since  the  rods  are  only  3  inches  from  the  neutral 
axis  £2  =  %X  10X100  =  600  pounds  per  square  inch, 
while  the  steel  may  safely  bear  twenty-five  times  as  great 
a  unit-stress.  The  full  safe  strength  of  the  steel  rods 
cannot  indeed  be  developed  unless  the  tensile  strength  of 
the  concrete  is  entirely  overcome.  A  symmetric  arrange- 
ment of  the  rods,  like  that  in  Fig.  52,  is  not  economical 
and  is  rarely  used  for  beams.  The  above  discussion  indi- 
cates, however,  the  principles  involved,  and  the  formulas 
will  be  modified  in  the  next  article  so  as  to  apply  to  the 
usual  cases  of  unsymmetric  reinforcement. 

Prob.  71.  A  beam  like  Fig.  52  is  10  inches  wide,  12  inches  deep, 
14  feet  in  span,  and  has  eight  steel  rods  each  %  inches  in  diameter, 
four  being  1  inch  below  the  top  and  four  being  1  inch  above  the 
bottom.  Compute  the  unit-stresses  for  the  1:2:4  concrete  and  for 
the  steel  when  the  beam  is  loaded  with  200  pounds  per  linear  foot 
besides  its  own  weight. 

ART.  72.    UNSYMMETRIC  REINFORCEMENT 

For  the  reasons  stated  in  the  last  article,  reinforced 
concrete  beams  are  generally  built  with  imbedded  rods 
only  on  the  tensile  side,  as  shown  in  Fig.  53.  Let  Si  be 
the  compressive  unit-stress  on  the  upper  surface  of  the 
beam,  T\  the  tensile  unit-stress  on  the  lower  surface,  and 
Ti  the  tensile  unit-stress  in  the  steel.  Let  n,  as  before, 
represent  the  ratio  E2/Ei  found  by  dividing  the  modulus 
of  elasticity  of  the  steel  by  that  of  the  concrete.  Let  6 
be  the  breadth  and  d  the  depth  of  the  rectangular  section, 
A  the  section  area  of  the  steel  rods,  the  centers  of  which 
are  at  the  distance  h  below  the  middle  of  the  beam. 
Let  M  be  the  bending  moment  of  the  loads  for  the  given 


138  REINFORCED    CONCRETE  CH.  9 

section,  and  let  it  be  required  to  find  the  values  of  Si, 
Ti,  and  Tz  due  to  M.  The  following  formulas  are  demon- 
strated in  Mechanics  of  Materials  (eleventh  edition). 

Case  I — When  the  loads  on  the  beam  are  light,  so  that 
the  unit-stress  on  the  tensile  side  does  not  exceed  about 
one-half  of  the  tensile  strength  of  the  concrete,  the  dis- 
tribution of  stresses  in  the  vertical  section  is  that  shown  in 

I---6— *i 


©   ©   © 


Fig.  53 

Fig.  53.  The  neutral  surface  in  this  case  lies  at  a  certain 
distance  g  below  the  middle  of  the  beam,  and  the  value 
of  g  may  be  computed  from 

h 


Then  the  unit-stresses  for  the  concrete  are 

d  +  2g       QM  d-2g       6M 

d*  +  I2gh  '    bd  d*  +  12gh  '   bd 

while  that  for  the  steel  is 

=  2n(h  -  g)     QM 
d*  +  12gh  '   bd 

and  from  these  formulas  the  beam  may  be  investigated, 
provided  the  load  does  not  produce  a  value  of  TI  greater 
than  about  100  pounds  per  square  inch. 

For  example,  let  6  =  10  and  d  =  12  inches,  h  =  4  inches, 
A  =  2.4  square  inches,  and  n  =  15  for  1:3:6  concrete 


ART.  72         TJNSYMMETRIC  REINFORCEMENT  139 

(Art.  68).  Then  g=  0.923  inches  is  the  distance  of  the 
neutral  surface  below  the  middle  of  the  beam.  Now  let 
the  span  of  the  beam  be  8  feet  and  the  uniform  load  upon 
it  be  2400  pounds;  the  bending  moment  at  the  middle 
is  M  =  1/%WI = 28  800  inch-pounds.  The  compressive  unit- 
stress  on  the  upper  surface  of  the  concrete  is  then  found 
to  be  Si  =  106  pounds  per  square  inch,  and  the  tensile 
unit-stress  on  the  lower  surface  is  Ti  =  7S  pounds  per 
square  inch,  while  the  tensile  unit-stress  in  the  steel  is 
Tz=l  110  pounds  per  square  inch.  Both  Si  and  Tz  are 
quite  small,  but  T  is  about  one-third  of  the  ultimate 
tensile  strength. 

Case  II — When  a  heavy  load  is  applied  to  a  reinforced 
concrete  beam  the  tensile  resistance  of  the  concrete  is  first 
overcome,  vertical  cracks  extending  upward  from  the  lower 
side,  and  thus  a  greater  stress  is  thrown  upon  the  steel. 
The  theoretic  analysis  for  this  case  is  a  difficult  one  unless 
it  is  assumed  that  the  concrete  below  the  neutral  surface 


L'f  r* 

£_.J 


"«2 

Fig.  54 

exerts  no  material  resistance,  and  this  assumption  is 
the  one  usually  made.  Fig.  54  shows  the  distribution 
of  stresses,  the  neutral  surface  being  usually  above  the 
middle  of  the  beam.  Let  6  be  the  breadth  of  the  beam, 
d  the  distance  of  the  centers  of  the  reinforcing  rods  below 
the  top,  and  kd  the  distance  of  the  neutral  surface  below 


140  REINFORCED   CONCRETE  CH.  9 

the  top,  k  being  a  number  less  than  unity.     Also  let  the 
ratio  A/bd  be  called  p.     First,  compute  k  from 

k=  -  np+  V2np  +  (np)2 
and  then  the  unit-stresses  are 

_2M_  M 

-  - 


the  first  being  for  compression  on  the  concrete  and  the 
second  for  tension  on  the  steel.  As  an  example,  let  6  =  12 
and  d  =  4.5  inches,  A  =  0.6  square  inches,  1  =  60  inches,  the 
uniform  load  be  1800  pounds,  and  n  =  10.  Then  k  =  0.373 
which  locates  the  depth  of  the  neutral  surface  below  the 
top  of  the  beam.  From  the  given  load  and  span,  M  is 
found  to  be  13,500  inch-pounds,  and  then  from  formula 
(11)  the  compressive  stress  on  the  upper  surface  of  the 
concrete  is  C  =  340  pounds  per  square  inch,  and  the  tensile 
stress  in  the  steel  rods  is  71  =  5710  pounds  per  square  inch. 
Both  of  these  are  lower  than  maximum  allowable  values. 
The  formulas  of  Case  II  are  those  usually  required  for 
the  investigation  of  a  reinforced  concrete  beam  unless  it 
is  so  lightly  loaded  that  the  unit-stress  T\,  found  by  the 
formula  of  Case  I,  is  less  than  about  100  pounds  per 
square  inch. 

Prob.  72.  Let  a  reinforced  beam  of  1:  2:  4  concrete  be  24  inches 
wide,  5  inches  deep,  and  6  feet  span,  with  1.2  square  inches  of  steel 
at  2  inches  below  the  middle;  compute  the  unit-stresses  S\,  Ti,  T*, 
due  to  the  light  uniform  load  of  1125  pounds.  Also  compute  the 
unit-stresses  C  and  T  due  to  the  heavy  uniform  load  of  4500  pounds. 

ART.  73.    DESIGN  OF  BEAMS 

When  a  reinforced  concrete  beam  is  to  be  designed 
the  allowable  unit-stresses  for  concrete  and  steel  are 


ART.  73  DESIGN  OF  BEAMS  141 

given  or  assumed,  as  also  the  span  and  width,  and  it 
is  then  required  to  compute  the  depth  of  the  beam  and 
the  section  area  of  the  steel.  When  this  is  done  according 
to  the  formulas  applicable  to  Fig.  53,  the  steel  is  stressed 
but  slightly;  if  T\  is  taken  as  100  pounds  per  square  inch 
for  the  concrete,  the  stress  T2  for  the  steel  will  be  less 
than  1500  pounds  per  square  inch.  It  is  found  impossible 
to  economically  design  a  beam  on  this  theory  and  have 
unit-stresses  prevail  that  are  satisfactory,  this  being  due 
to  the  low  tensile  strength  of  the  concrete.  Nothing 
remains  to  be  done,  therefore,  but  to  allow  the  concrete 
to  crack  on  the  tensile  side  and  thus  to  stress  the  steel 
higher  in  tension  than  is  otherwise  possible.  The  formulas 
(11)  given  above  for  the  distribution  of  stresses  shown  in 
Fig.  54  may  be  transformed  so  as  to  be  applicable  to  cases 
of  design. 

The  quantities  usually  given  in  designing  are  the 
allowable  compressive  unit-stress  C  on  the  concrete,  the 
allowable  tensile  unit-stress  T  on  the  steel,  the  ratio 
Ez/Ei  =  n,  the  bending  moment  M,  and  the  breadth  b 
of  the  rectangular  beam.  It  is  required  to  find  the  depth 
d  of  the  beam  and  the  section  area  A  of  the  reinforcing 
steel.  Let  the  ratio  T/C  be  designated  by  t,  then  for  the 
case  of  Fig.  54, 


are  the  formulas  for  computing  d  and  A.  The  unit-stress 
C  should  be  taken  as  high  as  allowable  by  the  specifica- 
tions; T  should  not  be  higher  than  the  highest  allowable 
value,  but  it  may  be  taken  lower  than  this  value  if  econ- 
omy in  cost  is  promoted. 


142  REINFORCED    CONCRETE  CH.    9 

For  example,  a  rectangular  beam  of  1  :  3  :  6  concrete 
is  to  have  a  span  of  14  feet,  a  breadth  of  20  inches,  and  is 
to  carry  a  uniform  load  of  300  pounds  per  square  foot, 
including  its  own  weight.  It  is  required  to  find  the  depth 
of  the  beam  and  the  section  area  of  the  reinforcing  rods 
so  that  the  unit-stresses  C  and  T  shall  be  350  and  14  000 
pounds  per  square  inch  respectively.  Here  ft  =15,  t  = 
T/C = 40,  and  6  =  20  inches.  The  total  load  on  the  beam  is 
300  X 14  X  20/12  =  7000  pounds,  and  the  bending  moment  is 
M  =  y8  X  7000  X 14  X 12  =  147  000  inch-pounds.  Inserting 
these  values  in  the  first  of  the  above  formulas,  there  is 
found  d=  13.0  inches.  Then  bd  =  20X  13  =  260  square  in- 
ches, and  from  the  second  formula  A  =0.90  square  inches. 

For  the  above  case  the  section  area  of  the  steel  is 
about  one-third  of  one  percent  of  the  section  area  bd 
of  the  concrete.  Higher  percentages  of  steel  are  fre- 
quently used,  from  0.60  to  1.25  percent  being  common 
values,  but  these  are  probably  not  economical  except 
for  high-class  concrete  and  low-priced  steel.  The  depth  d 
computed  by  the  above  formula  is  that  from  the  top  of  the 
beam  to  the  centers  of  the  reinforcing  rods.  The  actual 
depth  of  the  beam  is,  however,  greater  than  d  by  1  or  1^4 
inches,  the  extra  thickness  of  concrete  serving  to  protect 
the  steel  from  corrosion  and  from  the  effects  of  fire. 

Prob.  73  A.  For  the  above  numerical  data  except  that  for  T, 
compute  the  depth  d  and  the  section  A.  taking  the  value  of  T  as 
12  000  pounds  per  square  inch.  Also  taking  the  value  of  T  as  9000 
pounds  per  square  inch.  If  steel  costs  50  times  as  much  as  concrete, 
per  cubic  unit,  which  of  the  three  beams  is  the  cheapest? 

Prob.  73  B.  Design  two  reinforced  concrete  beams  12  inches 
wide  for  a  span  of  12  feet  6  inches  and  a  uniform  load  of  1500 
pounds  per  linear  foot,  one  being  1:2:4  concrete  and  the  other 
1:3:6  concrete. 


ART.  74  GENERAL  DISCUSSIONS  143 

ART.  74.    GENERAL  DISCUSSIONS 

The  formulas  and  methods  above  presented  for  rein- 
forced concrete  beams  are  valid  when  the  unit-stresses  in 
the  concrete  are  proportional  to  their  distances  from  the 
neutral  surface,  and  this  is  the  case  only  when  the  changes 
of  length  are  proportional  to  the  unit-stresses.  Concrete 
is  a  material  in  which  this  proportionality  does  not  exist 
for  compressive  stresses  higher  than  500  pounds  per 
square  inch,  so  that  it  cannot  be  expected  that  the 
formulas  will  apply  to  cases  of  rupture.  Formulas  for 
rupture  have  been  deduced  by  Hatt  and  others  in  which 
the  unit-stresses  are  taken  as  varying  according  to  a 
parabolic  law  with  their  distances  from  the  neutral 
surface,  and  such  formulas  are  sometimes  used  for  design- 
ing beams  by  applying  proper  factors  of  safety. 

The  phenomena  of  failure  of  reinforced  concrete  beams 
have  been  fully  ascertained  by  the  experiments  made  by 
Talbot  in  1905.  The  beams  were  tested  by  applying 
two  concentrated  loads  at  the  third  points  of  the  span, 
and  the  deflections  at  the  middle  were  measured  for 
several  increments  of  loading,  as  also  horizontal  changes 
of  length  near  the  top  and  bottom.  Under  light  loads 
the  tensile  resistance  of  the  concrete  was  plainly  apparent; 
when  the  tensile  unit-stress  in  the  concrete  approached 
the  ultimate  strength,  the  neutral  surface  rose  and  the 
stress  in  the  steel  increased.  A  little  later  fine  vertical 
cracks  appeared  on  the  tensile  side,  while  the  tensile 
stresses  in  the  steel  and  the  compressive  stresses  in  the 
concrete  increased  faster  than  the  increments  of  the  load. 
The  last  stage  was  a  rapid  increase  in  the  deformations, 
and  rupture  generally  occurred  by  the  crushing  of  the 


144  REINFORCED    CONCRETE  CH.  9 

concrete  on  the  upper  surface,  the  steel  being  then  stressed 
beyond  its  elastic  limit. 

In  some  cases,  especially  in  short  beams,  failure  is 
observed  to  occur  by  an  oblique  shearing  near  the  sup- 
ports, and  this  may  be  prevented  by  inclined  or  vertical 
reinforcing  rods.  In  designing  a  beam,  however,  it  is 
usually  not  necessary  to  make  computations  for  this  shear- 
ing stress  because  the  dimensions  obtained  by  the  flexure 
formulas  are  for  a  beam  which  is  to  carry  only  about  one- 
fifth  or  one-sixth  of  the  load  that  causes  rupture. 

The  above  discussions  give  only  an  introduction  to  the 
subject  of  reinforced  concrete.  Many  questions  in 
regard  to  beams  remain  yet  to  be  settled,  but  it  is  believed 
that  the  methods  above  given  are  fundamental  and  not 
liable  to  essential  change.  It  seems  likely  that  the  use  of 
combined  concrete  and  steel,  not  only  for  beams  and 
columns,  but  for  arches,  walls,  piers,  dams,  and  aqueducts, 
is  to  increase  rapidly  and  become  a  most  important  feature 
in  engineering  construction. 

Prob.  74.  A  plain  concrete  beam,  12  inches  wide,  13 K  inches 
deep,  and  14  feet  span,  broke  under  two  single  loads,  each  of  1300 
pounds  and  placed  at  the  third  points  of  the  span.  Compute  the 
modulus  of  rupture  by  the  common  flexure  formula.  If  this  beam 
has  one  percent  of  steel  placed  1J^  inches  above  the  lower  surface, 
compute  the  values  of  C  and  T. 

ART.  75.    REVIEW  PROBLEMS 

Prob.  75  A.  Find  the  safe  load  for  a  short  column  of  1  :  2  :  4 
concrete  which  is  2X3  feet  in  section  area. 

Prob.  75  B.  A  short  timber  column,  6X8  inches  in  section, 
has  two  steel  plates,  each  %X8  inches,  bolted  to  the  8-inch  sides. 
Compute  the  load  P  when  the  timber  is  stressed  to  750  pounds  per 
square  inch. 


ART.  75  REVIEW  PROBLEMS  145 

Prob.  75  C.  A  pier  of  1:3:6  concrete,  6  feet  in  diameter,  is 
surrounded  by  a  cast-iron  casing  1.15  inches  thick.  What  part 
of  the  total  load  is  carried  by  the  concrete? 

Prob.  75  D.  For  a  1  :  2  :  4  concrete  beam  like  Fig.  52  let  6  =  10 
and  d  =  12  inches,  and  the  six  steel  rods  be  5  inches  from  the  neutral 
surface.  Find  the  size  of  the  rods  when  the  beam  is  13  feet  in  span 
and  carries  a  total  uniform  load  of  5000  pounds. 

Prob.  75  E.  For  the  dimensions  in  Prob.  72,  what  uniform  load 
will  probably  cause  the  concrete  to  begin  to  fail  in  tension? 

Prob.  75  F.  Design  a  beam  of  1  :  2  :  4  concrete  with  three  rein- 
forcing rods,  the  width  to  be  8  inches  and  the  span  6  feet,  so  that  it 
will  safely  carry  a  total  uniform  load  of  2500  pounds. 

Prob.  75  G.  Consult  a  paper  by  Sewall  and  the  accompanying 
discussions  in  Vol.  56  of  Transactions  of  American  Society  of  Civil 
Engineers,  and  ascertain  different  opinions  as  to  what  should  be  the 
comparative  cost  of  concrete  and  steel  in  order  to  produce  the  most 
economical  reinforced  beam. 

Prob.  75  H.  If  a  force  of  235  pounds,  acting  at  the  end  of  a  lever 
17.5  inches  long,  twists  the  end  of  a  shaft  of  6.5  feet  length  through 
an  angle  of  14°  45',  what  force  acting  at  the  end  of  a  lever  9.75 
inches  long  will  cause  a  twist  of  28°  31'  when  the  length  of  the  shaft 
is  10.62  feet? 

Prob.  75  K.  What  center  load  will  rupture  a  wooden  beam  1  inch 
square  and  12  inches  in  span? 

Prob.  75  L.  A  wrought-iron  simple  beam  is  2X2  inches  in 
section.  What  must  be  its  length  so  that  it  will  rupture  under  its 
own  weight? 


146  COMBINED   STRESSES  CH.   10 


CHAPTER  10 

COMBINED   STRESSES 

ART.  76.     COMPRESSION  AND  FLEXURE 

When  a  simple  beam  is  subject  to  compressive  forces 
at  its  ends  the  compressive  stress  due  to  this  force  increases 
the  deflection,  and  hence  also  the  compressive  stress  on 
the  upper  side  of  the  beam.  Let  M  be  the  bending  moment 
at  the  middle  of  the  beam,  found  as  in  Art.  21,  c  the  dis- 
tance of  the  neutral  surface  of  the  beam  from  the  com- 
pression side  (Art.  23),  and  /  the  moment  of  inertia  of  the 
section  area  (Art.  24);  then  the  unit-stress  due  to  the 
simple  flexure  is  Me/ 1  (Art.  30).  Let  A  be  the  section 
area  of  the  beam  and  P  the  compressive  load  acting  on  the 
ends;  then  P/A  is  the  unit-stress  due  to  the  direct  com- 
pression (Art.  1).  Roughly  and  approximately,  then,  the 
total  compressive  unit-stress  on  the  upper  side  of  the  beam 
is  Mc/I  +  P/A. 

A  more  extended  discussion  will  show  (as  in  Mechanics 
of  Materials,  tenth  edition,  p.  256)  that  the  following 
formula  obtains  for  the  case  where  the  simple  beam  is 
loaded  at  the  middle: 


P  ,Mc/(        PP  \ 
**-A+T/  V    12EJJ 


where  E  is  the  modulus  of  elasticity  of  the  material  (Art. 
52).  For  a  uniform  load  on  the  beam  the  number  12  is 
to  be  replaced  by  9.6. 

For  example,  let  a  simple  wooden  beam  16  feet  long, 


ART.  77  TENSION  AND  FLEXURE  147 

10  inches  wide,  and  9  inches  deep  be  under  an  axial  com- 
pression of  40  000  pounds,  while  at  the  same  time  it 
carries  a  total  uniform  load  of  2000  pounds.  Here 
M=y8Wl=4SQQQ  pound-inches,  c=4.5  inches,  7  =  i/126cP 
=  607.5  inches4,  Z  =  96  inches,  P- 40  000  pounds,  E  = 
1  500  000  pounds  per  square  inch,  and  A  =  90  square  inches. 
Inserting  these  values  in  the  formula,  there  is  found 
S  =  444 +428  =  872  pounds  per  square  inch  as  the  flexural 
unit-stress  at  the  middle  of  the  beam  due  to  the  combined 
compression  and  flexure. 

Prob.  76.  A  simple  wooden  beam,  10  inches  wide  and  4  feet  long, 
carries  a  uniform  load  of  500  pounds  per  linear  foot  and  is  subjected 
to  a  longitudinal  compression  of  40  000  pounds.  Find  the  depth 
of  the  beam  so  that  the  maximum  compressive  unit-stress  may  be 
800  pounds  per  square  inch. 

ART.  77.    TENSION  AND  FLEXURE 

Let  a  beam  be  subject  to  flexure  by  transverse  loads  and 
also  to  a  tension  in  the  direction  of  its  length.  The  effect 
of  the  tension  is  to  decrease  the  deflection  of  the  beam 
and  hence  also  the  flexural  unit-stress.  The  formula  of 
the  last  article  can  be  applied  to  this  case  by  a  change 
in  one  of  the  signs.  Thus  for  the  case  of  a  single  load 
at  the  middle  of  the  simple  beam, 


and  for  a  uniform  load  on  the  beam  the  number  12  is 
to  be  changed  to  9.6. 

For  example,  take  a  steel  eye  bar,  18  feet  long,  1  inch 
wide,  and  8  inches  deep,  which  is  under  a  longitudinal 
tension  of  80  000  pounds.  The  weight  of  this  beam  is 


148  COMBINED    STRESSES  CH.  10 

490  pounds  and  M=12230  pound-inches.  Also  c  =  4 
inches,  7  =  42.67  inches4,  #  =  30  000  000  pounds  per  square 
inch,  P  =  80  000  pounds,  and  A  =8  square  inches.  Then 
the  formula  gives  S=  10  000+950=  10  950  pounds  per 
square  inch  due  to  the  combined  flexure  and  tension. 
By  using  the  approximate  method  noted  in  the  first  para- 
graph of  the  last  article,  there  is  found  S  =  10  000+1240  = 
11  240  pounds  per  square  inch. 

Prob.  77.  A  light  steel  I  beam,  4  inches  deep,  and  10  feet  long> 
has  a  load  of  650  pounds  at  the  middle  and  is  under  the  longitudinal 
tension  of  20  000  pounds.  Compute  the  flexural  unit-stress  due  to 
the  combined  loads. 

ART.  78.     SHEAR  AND  AXIAL  STRESS 

Let  a  bar  having  the  section  area  A  be  subjected  to  the 
longitudinal  tension  P,  and  at  the  same  time  to  a  shear  V 
at  right  angles  to  its  length.  The  axial  unit-stress  on 
the  section  area  is  P/A  which  will  be  designated  by  S, 
and  the  shearing  unit-stress  is  V/A  which  will  be  denoted 
by  S8.  These  two  direct  stresses  combine  to  produce 
tensile,  compressive,  and  shearing  unit-stresses  in  other 
directions.  The  following  formulas,  demonstrated  in 
Mechanics  of  Materials,  Art.  105  (tenth  edition),  give  the 
greatest  of  these  internal  unit  stresses: 

Max.  shearing  unit-stress,         $S'  =  \/ 
Max.  tensile  unit-stress,  Si  = 

Max.  compression  unit-stress,  $2  =      Ss'  —  VzS 
When  P  is  compression,  then  the  second  of  these  formulas 
gives  the  maximum  compressive  unit-stress  and  the  third 
gives  the  maximum  tensile  unit-stress. 

For  example,  take  a  bolt  one  inch  in  diameter  which  is 


ART.  79  COMPRESSION  AND  TORSION  149 

subject  to  a  longitudinal  tension  of  5000  pounds  and  at 
the  same  time  to  a  cross  shear  of  3000  pounds.  Here 
S  =  C366  and  S8  =  3820  pounds  per  square  inch.  Then  the 
first  formula  gives  $/  =  4970  pounds  per  square  inch  for 
the  maximum  shearing  unit-stress,  $i  =  8155  pounds  per 
square  inch  for  the  resultant  maximum  tensile  unit- 
stress,  and  82=  1790  pounds  per  square  inch  for  the  result- 
ant maximum  compressive  unit-stress  which  act  in  the  bolt 
in  directions  different  from  the  applied  unit-stresses. 

Prob.  78.  A  short  bolt  %  inch  in  diameter  is  subjected  to  a  longi- 
tudinal compression  of  2000  pounds  and  at  the  same  time  to  a 
cross  shear  of  3000  pounds.  Find  the  maximum  compressive,  ten- 
sile, and  shearing  unit-stresses  which  exist  in  the  bolt. 

ART.  79.     COMPRESSION  AND  TORSION 

Compression  and  torsion  are  combined  when  a  loaded 
vertical  shaft  rests  in  a  step  at  its  foot.  Here  there  is  a 
compressive  unit-stress  S  due  to  the  weight  of  the  shaft 
and  its  loads,  a  torsional  unit-stress  Ss  due  to  the  trans- 
mitted horse-power  (Art.  48).  These  combine  to  produce 
the  resultant  unit-stresses  Ssf,  Si,  and  S2  which  may  be 
computed  by  the  formulas  of  the  last  article. 

To  find  the  diameter  of  a  vertical  solid  shaft  for  this 
case  the  following  formula  may  be  used: 


V(16  Pp)2 

in  which  d  =  diameter,  $/  =  the  working  shearing  unit- 
stress,  Pp  =  the  twisting  moment  computed  by  Art.  48, 
W  =  the  weight  of  the  shaft  and  its  loads.  Assumed  values 
of  d  are  to  be  inserted  in  the  formula  until  one  is  found 
that  satisfies  it.  This  formula  also  serves  to  compute 
Sa'  directly  when  d  is  given. 


150  COMBINED    STRESSES  CH.  10 

A  vertical  shaft  is  sometimes  so  arranged  that  its  weight 
and  loads  are  supported  near  the  top  on  a  series  of  circular 
disks,  sometimes  called  a  thrust  bearing.  The  shaft  is 
thus  brought  into  tension  instead  of  compression,  and 
this  is  a  better  arrangement  because  there  is  then  no 
liability  to  flexure.  The  above  formula  applies  also  to 
this  case. 

Prob.  79.  A  vertical  shaft,  weighing  with  its  loads  6000  pounds, 
is  subjected  to  a  twisting  moment  by  a  force  of  300  pounds  acting 
at  a  distance  of  48  inches  from  its  center.  If  the  shaft  is  structural 
steel,  4  feet  long  and  2  inches  in  diameter,  find  its  factor  of  safety. 

ART.  80.     FLEXURE  AND  TORSION 

This  case  occurs  when  a  horizontal  shaft  for  the  trans- 
mission of  power  is  loaded  transversely  with  weights. 
Let  S  be  the  flexural  unit-stress  computed  from  (4)  and 
Ss  the  torsional  shearing  unit-stress  found  from  Art.  49. 
Then  by  the  last  article  the  resultant  maximum  unit- 
stresses  are 


Max.  shearing  unit-stress,    Ssf  =  \^S82  +  %=  S2 


Max.  flexural  unit-stress,    S'  =y£  +  VSS2  +  V±S2 

These  can  be  used  to  find  the  greatest  internal  unit-stress, 
and  then  the  factors  of  safety  of  the  material  are  known. 

It  is  thus  seen  that  the  actual  maximum  unit-stresses 
in  a  shaft  due  to  combined  flexure  and  torsion  are  much 
higher  than  those  derived  from  the  formulas  for  flexure 
or  torsion  alone.  In  determining  the  diameter  of  a  shaft 
it  is  hence  necessary  to  take  Ssf  as  the  allowable  shearing 
unit-stress  and  Sf  as  the  allowable  tensile  or  compressive 
unit-stress.  For  a  round  shaft  of  diameter  d,  the  value  of 


ART.  81  TENSION  AND  COMPRESSION  151 

S8  is  Ppc/J  where  Pp  is  the  twisting  moment  (Art.  45) 
or  Ss  =  IQPp/irtf,  while  the  value  of  S  is  Me/ 1  (Art.  30) 
or  S  =  32M/ird?.  Inserting  these  in  the  above  formulas 
they  reduce  to 


the  first  being  for  the  resultant  shearing  and  the  second 
for  the  resultant  tension  or  compression.  Since  the 
allowable  Sg'  is  smaller  than  the  allowable  S',  it  often 
happens  that  the  first  formula  will  give  a  larger  diameter 
than  the  second. 

As  an  example,  find  the  diameter  of  a  horizontal  steel 
shaft  to  transmit  90  horse-powers  at  250  revolutions 
per  minute,  when  the  distance  between  bearings  is  8  feet 
and  there  is  a  load  of  480  pounds  at  the  middle,  the 
allowable  unit-stress  being  Ssf  =  5000  and  Sf  =  7000  pounds 
per  square  inch.  Here  the  twisting  moment  is  Pp  = 
63030X90/250  =  22690  pound-inches,  and  the  bending 
moment  is  AT  =  480X96/8  =  5760  pound-inches.  Then 
using  the  first  formula  the  diameter  d  is  found  to  be  2.9 
inches,  while  from  the  second  formula  it  is  2.8  inches. 
Hence  the  shaft  should  be  about  3  inches  in  diameter. 

Prob.  80.  A  horizontal  steel  shaft  o"  17  inches  outer  and  11  inches 
inner  diameter  is  to  transmit  16  000  horse-powers  at  50  revolutions 
per  minute,  the  distance  between  bearings  being  18  feet.  Taking 
into  account  the  flexure  due  to  the  weight  of  the  shaft,  compute  the 
maximum  unit-stresses. 

ART.  81.     TENSION  AND  COMPRESSION 

When  a  tensile  unit-stress  Si  exists  in  a  body  acting 
in  a  certain  direction  and  a  second  tensile  unit-stress  S2 
is  applied  in  the  same  direction,  then  the  resultant  unit- 


152  COMBINED    STRESSES  CH.  10 

stress  is  Si+S2.  When  Si  is  tensile  and  S2  compressive 
then  the  resultant  unit-stress  is  Si  —  S2  ;  if  Si  is  greater  than 
£2  then  Si—  S2  is  tensile,  but  if  Si  is  less  than  £2  then  it 
is  compressive. 

When  a  tensile  unit-stress  Si  exists  in  a  body  acting  in  a 
certain  direction  and  a  second  tensile  unit-stress  S2  is 
applied  in  a  direction  at  right  angles  to  Si,  then  the  true 
resultant  unit-stress  T\  is  Si—\Sz  where  X  is  the  'factor 
of  lateral  contraction.'  The  mean  value  of  X  for  wrought 
iron  and  steel  is  about  %  while  for  cast  iron  it  is  about  J^. 
Thus  if  a  tensile  unit-stress  of  5000  pounds  per  square 
inch  acts  in  a  certain  direction  on  a  steel  bar  and  another 
tensile  unit-stress  of  6000  pounds  per  square  inch  acts 
at  right  angles  to  the  first,  then  the  true  resultant  tensile 
unit-stress  in  the  original  direction  is  T\  =  5000  —  %  (6000) 
=  3000  pounds  per  square  inch. 

Let  a  body  be  subject  to  tensile  forces  acting  in  three 
rectangular  directions,  as  for  instance  upon  the  faces  of  a 
cube.  Let  Si,  S*,  S3  be  the  tensile  unit-stresses  in  the 
three  directions.  Then  the  true  unit-stresses  acting  in 
these  three  directions  are 


If  any  unit-stress  S  is  compression  it  is  to  be  taken  as 
negative  in  the  formulas,  then  the  true  unit-stresses  are 
tensile  or  compressive  according  as  their  numerical  values 
are  positive  or  negative. 

As  an  example,  let  a  steel  bar  2  feet  long  and  3X2 
inches  in  section  area  be  subject  to  a  tension  of  60  000 
pounds  in  the  direction  of  its  length,  and  to  a  com- 


ART.  82  REVIEW  PROBLEMS  153 

pression  of  432  000  pounds  upon  the  two  flat  sides. 
Here  Si  =  60  000/6  =  10  000  pounds  per  square  inch,  S2  = 
—432  000/72  =  —  6000  pounds  per  square  inch,  and  S3  =  0. 
Then,  taking  X  as  1/3,  the  true  internal  unit-stresses  are 
Ti=  +12  000,  Tz=  -9330,  T3=  -1330  pounds  per  square 
inch.  For  this  case  the  true  tensile  unit-stress  Tt  is 
20  percent  greater  than  Si  and  the  true  compressive  unit- 
stress  T2  is  more  than  50  percent  greater  than  S2. 


Prob.  81.  A  common  brick,  2J^X4X8M  inches,  is  subject  to  a 
compression  of  3200  pounds  upon  it?  top  and  bottom  faces,  500 
pounds  upon  its  sides,  and  60  pounds  upon  its  ends.  Taking  X  as 
%,  compute  the  true  internal  unit-stresses  in  the  three  directions. 

ART.  82.     REVIEW  PROBLEMS 

Prob.  82  A.  An  I  beam,  12  inches  deep  and  weighing  35  pounds 
per  foot,  acts  as  a  simple  beam  with  a  span  of  30  feet.  Compute  the 
flexural  unit-stress  at  the  middle  due  to  its  own  weight. 

Prob.  82  B.  When  an  axial  compression  of  60  000  pounds  acts 
on  the  beam  of  the  last  problem,  find  the  unit-stress  due  to  the 
combined  compression  and  flexure. 

Prob.  82  C.  A  horizontal  eye  bar,  1J^X9  inches  in  section,  is 
under  a  tension  of  120  000  pounds.  Find  the  tensile  unit-stress  at 
the  middle  due  to  the  combined  tension  and  flexure. 

Prob.  82  D.  Compute  the  greatest  tensile,  compressive,  and  shear- 
ing unit-stresses  due  to  the  combination  of  a  direct  tension  of  24  000 
pounds  with  a  cross-shear  of  7500  pounds,  both  acting  on  a  bar 
1%  inches  in  diameter. 

Prob.  82  E.  A  vertical  steel  shaft  weighs  with  its  loads  64  000 
pounds  and  transmits  1200  horse-powers  at  75  revolutions  per 
minute.  What  should  be  the  diameter? 

Prob.  82  F.  A  cast-iron  ball  is  subjected  in  every  direction  to  a 
uniform  hydrostatic  pressure  of  625  pounds  per  square  inch.  What 
is  the  actual  true  compressive  unit-stress  which  exists  at  every  point 
within  the  ball? 


154  EESILIENCE    OF   MATERIALS  CH.  11 


CHAPTER  11 

RESILIENCE  OF   MATERIALS 
ART.  83.    FUNDAMENTAL  IDEAS 

When  a  force  of  uniform  intensity  P  is  exerted  through 
the  distance  e  the  work  performed  is  measured  by  the 
product  Pe.  When  a  bar  is  tested  in  a  machine,  however, 
the  force  gradually  and  uniformly  increases  from  0  up 
to  the  value  P  and  produces  the  elongation  e;  here  the 
work  performed  is  i/^Pe,  because  the  average  value  of  the 
uniformly  increasing  force  is  i/>P.  In  the  first  place  the 
work  may  be  represented  by  a  rectangle  of  height  P 
and  base  e;  in  the  second  case  the  work  may  be  repre- 
sented by  a  triangle  of  height  P  and  base  e;  and  the  area 
of  the  triangle  is  one-half  that  of  the  rectangle. 

As  the  external  force  increases  from  0  up  to  P  the 
internal  stress  in  the  bar  increases  gradually  and  uni- 
formly from  0  up  to  S.  The  internal  work  of  these 
stresses  is  called  the  'resilience'  of  the  bar.  As  the 
internal  work  equals  the  external  work  y2Pe,  this  quantity 
is  a  measure  of  the  resilience. 

Strength  is  the  capacity  of  a  body  to  resist  force; 
stiffness  is  the  capacity  of  a  body  to  resist  deformation; 
resilience  is  the  capacity  of  a  body  to  resist  work.  The 
higher  the  resilience  of  a  material  the  greater  is  its  capacity 
to  resist  the  work  of  external  forces. 

Elastic  resilience  is  that  internal  work  which  has  been 
performed  when  the  internal  stress  reaches  the  elastic 


ART.  84          ELASTIC  RESILIENCE  OF  BARS  155 

limit.  Ultimate  resilience  is  that  internal  work  which 
has  been  performed  when  the  body  is  ruptured.  Ultimate 
strength  is  usually  from  two  to  three  times  the  elastic 
strength;  ultimate  elongation  is  always  much  greater 
than  elastic  elongation;  and  ultimate  resilience  is  very 
much  larger  than  elastic  resilience. 

Resilience,  like  work,  is  expressed  in  foot-pounds,  or 
inch-pounds,  usually  in  the  latter  unit.  Thus,  if  a  bar 
is  subject  to  a  stress  which  gradually  and  uniformly 
increases  from  0  up  to  5000  pounds  and  is  accompanied 
by  an  elongation  of  0.5  inches,  the  resilience  is  1250 
inch-pounds. 

Prob.  83.  A  wrought-iron  bar  weighing  30  pounds  per  linear  foot 
is  subject  to  a  stress  of  5000  pounds  per  square  inch  which  is  accom- 
panied by  an  elongation  of  0.25  inches.  What  is  the  resilience  in 
inch-pounds? 

ART.  84.     ELASTIC  RESILIENCE  OF  BARS 

Let  a  bar  of  length  Z  and  section  area  A  be  under  a 
tension  P,  which  produces  a  unit-stress  S  equal  to  the 
elastic  limit  of  the  material  and  an  elongation  e.  The 
elastic  resilience  of  the  bar  is  then  equal  to  ^Pe.  Now 
P  =  SA,  and  by  Art.  53  the  elastic  elongation  is  e  = 
Pl/AE  =  Sl/E;  hence  letting  K  represent  the  elastic 
resilience,  the  product  y^Pe  becomes 


K=JEAl  <12) 

or  the  elastic  resilience  of  a  bar  is  proportional  to  its 
section  area  and  to  its  length,  that  is,  to  its  volume. 

When  the  bar  has  a  section  of  one  square  inch  and  a 
length  of  one  inch,  then  Al  is  one  cubic  inch,  and  the 


156  RESILIENCE    OF   MATERIALS  CH.  11 

elastic  resilience  is 

S* 
''~2E 

in  which  S  is  the  elastic  limit  of  the  material. 

This  quantity  k  is  called  the  modulus  of  resilience,  since 
for  any  given  material  it  is  a  constant.  For  bars  under 
tension  the  average  values  of  S  are  given  in  Art.  2,  and 
those  of  E  in  Art.  52.  Using  these  constants,  the  'mod- 
ulus of  resilience'  k  has  the  following  values  for  tension: 

For  timber,  k  =  3  inch-pounds  ^ 

For  cast  iron,          k=   1  inch-pound   V 
For  wrought  iron,  k=l2  inch-pounds  V 

For  hard  steel,        k  =  42  inch-pounds    * 

t 

These  figures  show  that  the  capacity  of  steel  to  resist 
work  within  the  elastic  limit  is  the  greatest  of  the  four 
materials,  and  that  of  cast  iron  the  least. 

For  a  bar  of  any  size  the  elastic  resilience  is  found 
by  multiplying  its  volume  by  the  modulus  of  resilience  k, 
Thus,  a  bar  of  timber  whose  volume  is  50  cubic  inches 
has  an  elastic  resilience  of  about  150  inch-pounds,  that  is, 
the  external  work  required  to  stress  it  up  to  the  elastic 
limit  is  150  inch-pounds.  The  particular  shape  of  the 
bar  is  unimportant;  it  may  be  5  inches  in  section  area  and 
10  inches  long,  or  2  inches  in  section  area  and  25  inches 
long,  or  any  other  dimensions  which  give  a  volume  of  50 
cubic  inches. 

The  above  formula  (12)  also  gives  the  work  required 
to  produce  any  unit-stress  S  which  is  less  than  the  elastic 
limit.  For  example,  let  it  be  required  to  find  the  work 
needed  to  stress  a  bar  of  wrought  iron  up  to  12  500  pounds 


ART.  85         ELASTIC  RESILIENCE  OF  BEAMS  157 

per  square  inch,  the  diameter  of  the  bar  being  2  inches 
and   its   length    18   feet.      Here  £=12500   pounds   per 
square    inch,    #  =  25000000    pounds    per    square    inch, 
J.=3.14  square  inches,  and  Z  =  216  inches.    Then 
12  5002X3.  14X216     ( 


2X25000000 

If  the  bar  is  required  to  undergo  this  stress  250  times  per 
minute,  the  work  required  in  one  minute  is  250X2120  = 
530000  inch-pounds  =  44  200  foot-pounds.  The  power 
expended  in  stressing  the  bar  is  hence  44  200/33  000  =  1.34 
horse-powers. 

When  a  bar  is  under  a  unit-stress  Si  and  this  is  in- 
creased by  additional  exterior  loads  to  $2,  the  resilience 
due  to  these  loads  is 


provided  *S>2  be  not  greater  than  the  elastic  limit. 

Prob.  84.  A  bar  of  steel  10  feet  long  and  weighing  490  pounds 
is  stressed  in  one  second  from  4000  up  to  9000  pounds  per  square 
inch.  What  work  and  what  horse-power  are  expended  in  doing  this? 

ART.  85.     ELASTIC  RESILIENCE  OF  BEAMS 

When  a  simple  beam  of  span  I  is  brought  into  stress 
by  a  load  P  applied  gradually  and  uniformly  at  the  middle, 
the  deflection  /  results  and  the  work  y2Pf  is  performed. 
This  work  equals  the  resilience  of  the  beam.  The  value 
of  /  in  terms  of  the  horizontal  unit-stress  is  given  in  Art.  56, 
and  the  value  of  P  in  terms  of  the  unit-stress  S  is  given 
by  (4)  of  Art.  28.  Accordingly  the  product  %-P/"  has  the 
value 


158  RESILIENCE    OF   MATERIALS  CH.  11 

in  which  7,  the  moment  of  inertia  of  the  section,  has 
been  replaced  by  its  equivalent  Ar*,  where  A  is  the  section 
area  and  r  its  least  radius  of  gyration  (Art.  37). 

For  a  simple  beam  under  a  full  uniform  load  the  elastic 
resilience  is  given  by 


which  is  1%  times  that  of  the  simple  beam  with  a  single 
load  at  the  middle. 

These  expressions  show  that  the  elastic  resilience  of 
beams  of  similar  cross-sections  is  proportional  to  their 
volumes.  For  rectangular  sections  where  the  depth  is 
d,  the  value  of  c  is  y^d  and  that  of  r2  is  i/fad2;  thus  r2/c2 
is  y$.  Hence  a  rectangular  bar  under  tensile  stress  has 
nine  times  the  resilience  of  a  rectangular  beam  loaded 
at  the  middle  and  5%  times  that  of  the  same  beam  under 
a  full  uniform  load. 

Prob.  85.  What  horse-power  is  required  to  stress  in  one  second 
a  heavy  20-inch  steel  I  beam  of  24  feet  span  from  500  up  to  8000 
pounds  per  square  inch,  this  being  done  by  a  load  at  the  middle? 

ART.  86.     ULTIMATE  RESILIENCE 

The  ultimate  resilience  of  a  body  is  equal  to  the  external 
work  required  to  produce  rupture.  The  ultimate  resilience 
greatly  surpasses  the  elastic  resilience,  it  being  for  wrought 
iron  and  steel  sometimes  five  hundred  times  as  large. 
It  is  not  possible,  however,  to  establish  a  formula  by 
which  the  ultimate  resilience  can  be  computed,  because 
the  law  of  increase  of  the  deformations  beyond  the  elastic 
limit  is  unknown. 

If  a  diagram  is  made  showing  the  increase  of  elonga- 


ART.  87  SUDDEN  LOADS  159 

tion  with  stress,  as  in  the  figure  of  Art.  4,  the  abscissas 
indicating  the  elongations  and  the  ordinates  the  stresses, 
then  the  area  included  between  the  curve  and  the  axis 
of  elongations  represents  the  ultimate  resilience  for  one 
cubic  inch  of  the  material.  The  total  ultimate  resilience 
is  then  found  by  multiplying  this  area  by  the  volume 
of  the  specimen  in  cubic  inches. 

In  Art.  15  it  was  remarked  that  the  product  of  the 
ultimate  strength  and  ultimate  elongation  is  an  index  of 
the  quality  of  wrought  iron  and  steel.  This  is  so  because 
it  is  a  rough  measure  of  the  ultimate  resilience  or  resistance 
to  external  work.  A  measure  which  more  closely  fits  the 
area  given  by  a  stress-diagram  is 


in  which  Se  is  the  elastic  limit,  St  the  ultimate  tensile 
strength,  and  s  the  ultimate  unit-elongation.  For  example, 
take  a  wrought-iron  specimen  where  Se  =  25  000  and  St  = 
50  000  pounds  per  square  inch,  while  s  =  30  percent  =  0.30  ; 
then  k  =  12  500  inch-pounds  is  the  ultimate  resilience  for 
one  cubic  inch  of  the  material. 

Prob.  86.  Show  from  the  values  given  in  Arts.  2  and  4  that  the 
average  ultimate  resilience  of  timber  in  tension  is  about  50  percent 
greater  than  that  of  cast  iron. 

ART.  87.  SUDDEN  LOADS 

When  a  tension  is  gradually  applied  to  a  bar  it  increases 
from  0  up  to  its  final  value,  while  the  elongation  increases 
from  0  to  e  and  the  unit-stress  increases  from  0  to  S.  A 
'sudden  load'  is  one  which  has  the  same  intensity  from 
the  beginning  to  the  end  of  the  elongation;  this  elonga- 
tion being  produced,  the  bar  springs  back,  carrying  the 


160  EESILIENCE    OF   MATERIALS  CH.  11 

load  with  it,  and  a  series  of  oscillations  results,  until 
finally  the  bar  comes  to  rest  with  the  elongation  e.  The 
temporary  elongation  produced  is  greater  than  e,  and 
hence  also  the  temporary  stresses  produced  are  greater 
than  S. 

Let  P  be  a  suddenly  applied  load  and  y  the  temporary 
elongation  produced  by  it;  the  external  work  performed 
during  its  application  is  Py.  Now  let  Q  be  the  internal 
stress  corresponding  to  the  elongation  y;  this  increases 
gradually  and  uniformly  from  0  up  to  Q,  and  hence  its 
resilience  or  internal  work  is  ^/^Qy-  But,  since  internal 
work  must  equal  external  work, 

Py,     or     Q  =  2P 

that  is,  the  sudden  load  P  produces  a  temporary  internal 
stress  equal  to  2P. 

Now  after  the  oscillations  have  ceased,  the  bar  comes 
to  rest  under  the  steady  load  P  and  has  the  elongation  e. 
If  the  elastic  limit  of  the  material  has  not  been  exceeded, 
corresponding  elongations  are  proportional  to  their 
stresses;  thus 


that  is,  the  sudden  load  produces  a  temporary  elongation 
double  that  caused  by  the  same  load  when  gradually 
applied. 

If  A  is  the  section  area  of  the  bar  the  unit-stress  S 
under  the  gradual  load  is  P/A,  and  the  temporary  unit- 
stress  produced  under  the  sudden  load  is  2P/A  or  2S. 
The  unit-stresses  temporarily  produced  by  sudden  loads 
are  hence  double  those  caused  by  steady  loads.  It  is  for 


ART.  88  STRESSES  DUE  TO  IMPACT  161 

this  reason  that  factors  of  safety  are  taken  higher  for 
variable  loads  then  for  steady  ones. 

Prob.  87.  A  simple  beam  of  wrought  iron,  2X2  inches  and  18 
inches  long,  is  1o  be  loaded  with  3000  pounds  at  the  middle.  Show 
that  the  beam  will  be  unsafe  if  this  be  applied  suddenly. 

ART.  88.     STRESSES  DUE  TO  IMPACT 

Impact  is  said  to  be  produced  in  a  bar  or  beam  when 
a  load  falls  upon  it  from  a  certain  height.  The  temporary 
stresses  and  deformations  in  such  a  case  are  greater 
than  for  sudden  loads,  and  may  often  prove  very  injurious 
to  the  material.  If  the  elastic  limit  is  not  exceeded,  it  is 
possible  to  deduce  an  expression  showing  the  laws  that 
govern  the  stresses  produced  by  the  impact.  This  will 
here  be  done  only  for  the  case  of  impact  on  the  end  of  a  bar. 

When  the  load  P  falls  from  the  height  h  upon  the  end 
of  a  bar  and  produces  the  momentary  elongation  y,  the 
work  performed  is  P(/i+?/)  .  The  stress  in  the  bar  increases 
gradually  and  uniformly  from  0  up  to  the  value  Q,  so 
that  the  resilience  or  internal  work  is  y^Qy.  Hence  there 
results 


Also,  if  e  is  the  elongation  due  to  the  static  load  P,  the 
law  of  proportionality  of  elongation  to  stress  gives 

y=Q 

e     P 
By  solving  these  equations  the  values  of  Q  and  y  are 


162  RESILIENCE    OF   MATERIALS  CH.  11 

which  give  the  temporary  stress  and  elongation  produced 
by  the  impact. 

If  h  =  0  these  formulas  reduce  to  Q  =  2P  and  y  =  2e, 
as  found  in  the  last  article  for  sudden  loads.  If  h  =  4e 
they  become  Q  =  4P  and  ?/  =  4e;  if  h=l2e  they  give 
Q  =  GP  and  y  —  6e.  Since  e  is  a  small  quantity  for  any 
bar  it  follows  that  a  load  P  dropping  from  a  moderate 
height  upon  the  end  of  a  bar  may  produce  great  tem- 
porary stresses  and  elongations.  If  these  stresses  exceed 
the  elastic  limit  they  cause  molecular  changes  which 
result  in  brittleness  and  render  the  material  unsafe. 

The  above  expressions  for  Q  and  y  are  not  exact,  as 
the  resistance  against  motion  due  to  the  inertia  of  the 
material  has  not  been  taken  into  account.  In  Chapters 
XIII  and  XIV  of  Mechanics  of  Materials  (tenth  edition) 
the  subject  is  discussed  far  more  completely  than  has  been 
possible  here. 

Prob.  88.  In  an  experiment  upon  a  spring  a  steady  weight  of 
15  ounces  on  the  end  produced  an  elongation  of  0.4  inches.  What 
temporary  elongation  would  be  produced  when  the  same  weight 
is  dropped  upon  the  end  of  the  spring  from  a  height  of  7  inches? 

ART.  89.     REPEATED  STRESSES 

Ultimate  strength  is  usually  understood  to  be  that 
steady  unit-stress  which  causes  the  rupture  of  a  bar  in 
one  application.  Experience  and  experiment  teach, 
however,  that  rupture  may  be  caused  by  a  unit-stress  less 
than  the  ultimate  strength  when  that  unit-stress  is  applied 
to  a  bar  a  large  number  of  times  in  succession.  For 
example,  Wohler  showed  that  a  bar  of  wrought  iron  could 
be  broken  in  tension  by  800  applications  of  52  800  pounds 


ART.  89  REPEATED  STRESSES  163 

per  square  inch  and  by  10  140  000  applications  of  35  000 
pounds  per  square  inch,  the  range  of  stress  in  each  applica- 
tion being  from  0  up  to  the  designated  value. 

It  has  also  been  shown  by  Wohler  and  others  that  the 
greater  the  range  of  stress,  the  less  is  the  unit-stress 
required  to  rupture  it  with  a  large  number  of  applications. 
Also  that  when  the  range  of  unit-stress  is  from  0  up  to 
the  elastic  limit,  rupture  occurs  only  after  an  enormous 
number  of  applications. 

Let  PI  be  the  least  and  P2  the  greatest  tensile  stresses 
which  occur  in  a  bar  under  repeated  stress  and  let  n  be 
the  ratio  Pi/P2.  Let  Su  be  the  ultimate  strength  and  Se 
the  elastic  limit  of  the  material.  Then  Weyrauch's  for- 
mula for  the  unit-stress  which  ruptures  the  bar  after  an 
enormous  number  of  repetitions  is 


For  structural  steel,  using  the  mean  values  given  in 
Tables  1  and  2,  this  becomes  £  =  35  000(1  +%ri)  and  for 
wrought  iron  $  =  25  000(1 +1/:jw)-  For  example,  let  a  bar 
of  structural  steel  range  in  tension  from  80  000  to  160  000 
pounds;  then  n  =  y2  and  £  =  47500  pounds  per  square 
inch  is  the  tensile  unit-stress  which  will  rupture  it  after 
an  enormous  number  of  applications,  although  the  ulti- 
mate strength  observed  in  one  application  is  65  000  pounds 
per  square  inch.  The  above  formula  also  applies  to  the 
case  where  PI  and  P2  are  both  compressive  stresses. 

When  PI  and  P2  are  stresses  of  different  kinds,  one 
being  tension  and  the  other  compression,  let  n  be  the  ratio 
Pi/Ps  without  respect  to  sign.  Then  Weyrauch's  formula 
for  the  unit-stress  which  produces  rupture  after  an 


164  RESILIENCE    OF   MATERIALS  CH.  11 

enormous  number  of  applications  is 


For  structural  steel  this  becomes  S  =  35  000(1  —  y^n)  and 
for  wrought  iron  S  =  25  000(1  —  y^ri).  For  example,  if 
the  forces  in  a  bar  of  structural  steel  range  from  80  000 
pounds  compression  to  160  000  pounds  tension,  then 
n  =  i/>  and  S  =  26  200  pounds  per  square  inch  is  the 
compressive  unit-stress  which  will  rupture  the  bar  after 
an  enormous  number  of  applications. 

The  above  formulas  are  sometimes  used  to  find  working 
unit-stresses  for  designing  bars,  such  a  factor  of  safety 
being  used  that  the  value  of  S  shall  be  less  than  one- 
half  of  Se. 

Prob.  89.  A  wrought-iron  bar  is  subject  to  axial  forces  which 
range  from  320  000  to  400  000  pounds.  Compute  the  rupturing 
unit-stress  S  after  an  enormous  number  of  repetitions,  first  when 
both  forces  are  tension,  second  when  the  smaller  one  is  compression 
and  the  larger  one  is  tension. 

ART.  90.    REVIEW  PROBLEMS 

Prob.  90  A.  How  many  foot-pounds  of  work  are  required  to 
stress  a  steel  piston-rod,  3  inches  in  diameter  and  4  feet  long,  from 
0  up  to  16  000  pounds  per  square  inch? 

Prob.  90  B.  What  horse-power  is  required  to  stress  the  rod  of 
the  last  problem  120  times  in  one  minute? 

Prob.  90  C.  Compute  the  horse-power  required  to  deflect,  59 
times  per  second,  a  wrought-iron  cantilever  beam,  2X3X72  inches, 
so  that  at  each  deflection  the  unit-stress  S  shall  range  from  0  to 
9000  pounds  per  square  inch. 

Prob.  90  D.  What  work  is  required  to  rupture  by  tension  a 
wrought-iron  bar  which  weighs  325  pounds? 

Prob.  90  E.    Discuss  the  case  where  a  sudden  load  P  is  applied 


ART.  91  ANSWERS  TO  PROBLEMS  165 

to  a  bar  which  is  already  under  the  static  load  PI.     What  is  the 
maximum  stress? 

Prob.  90  F.  What  is.  the  height  from  which  a  weight  must  fall 
upon  the  end  of  a  bar  in  order  to  produce  a  deformation  equal  to 
three  times  the  static  deformation  ? 

Prob.  90  G.  Consult  Test  of  Metals,  published  annually  by  the 
U.  S.  Ordnance  Department,  and  describe  some  of  the  endurance 
tests  made  by  Howard  on  rotating  shafts. 

Prob.  90  H.  A  bar  of  wrought  iron  is  to  be  subjected  to  re- 
peated stresses  ranging  from  16  000  pounds  tension  to  80  000 
pounds  tension.  What  should  be  its  diameter  using  a  safety  factor 
of  4  in  the  above  formula  for  S  ? 

Prob.  90  J.  Solve  the  last  problem,  taking  the  smaller  stress  as 
compression  and  the  larger  one  as  tension. 


ART.  91.     ANSWERS  TO  PROBLEMS 

Below  are  given  answers  to  a  few  of  the  problems 
stated  in  the  preceding  pages,  the  number  of  the  problem 
being  placed  in  parentheses.  However  satisfactory  it 
may  be  to  a  student  to  know  the  true  result  of  a  solution, 
let  him  remember  that  after  commencement  day  answers 
to  problems  will  never  be  given. 

(1  A)  4%2  inches.  (16  B)  3ys  inches. 

(3  C)  About  1  000  000  pounds.  (17  B)  105  000  pounds. 

(4  A)  4K  inches  for  wrought-  (ISA)  1200  pounds. 

iron  bar.  (23  B)  4.04  inches. 

(5  A)  1780  feet.  (24)  2097  inches4. 

(6  B)  Factors  are  6  and  16.  (25  J)  2840  pounds. 

(7  A)  3y8    inches    for    second  (29  A)  3130  pounds. 

case.  (31  A)  18  inches. 

(10  B)  Reduction  of  area  =  50.5  (33  A}  5.9  and  3.9. 

percent.  (35  D)  Nearly  8. 

(13  B)  $1058.84.  (35  E)  24  inches. 

(14  A)  66  000  pounds.  (35  F}  2%iol. 

(16  A)  4.04  cents.  (36)  54  000  pounds. 


166 


RESILIENCE    OF   MATERIALS 


CH.  11 


(39  A}  58  000  pounds. 

(665)   1500  pounds. 

(44  D)  31A. 

(67  D)  500  pounds  per  square 

(4AE)  13M  inches. 

inch. 

(45  A)  500  pounds. 

(74)       Sr  =  200  pounds  per 

(47  A)  1.3  inches. 

square  inch. 

(51  A)  30  pounds. 

(75  B)  126  000  pounds. 

(51  E)  2%  inches. 

(78)      Si  =  9420  pounds  per 

(51  F)  144  to  100. 

square  inch. 

(53  A)  0.12  inches. 

(79)       Nearly  6. 

(54  A)  0.0011  inches. 

(83)       5625  inch-pounds. 

(59  F)  14  500  000  pounds  per 

(88)       2.8  inches. 

square  inch. 

(90  F)  h  =  1.5  e. 

(63  A)  59  500  pounds. 

(90  H)  d  =  3.7  inches. 

Prob.  91.  The  horizontal  flexural  unit-stress  on  the  upper  sur- 
face of  a  railroad  rail  is  9500  pounds  per  square  inch  and  the  vertical 
compressive  unit-stress  due  to  the  weight  of  the  driving-wheel  is 
17  500  pounds  per  square  inch.  Compute  the  true  resultant  unit- 
stresses  in  the  horizontal  and  vertical  directions. 


INDEX 


Absorption  test,  24 
Aluminium,  35 
Angle  sections,  49 
Answers  to  problems,  165 

Bars,  4,  7,  29,  106,  128 

Beams,  37-76,  108,  124,  146,  157 
cantilever,  56,  59,  63,  68, 

107 

overhanging,  70 
reinforced,  134,  137,  140 
restrained,  110 
simple,  39,  56,  59,  68 
uniform  strength,  74 

Bending  moments,  41,  61,  63 

Bending  of  beams,  61,  69,  107 
columns,  77-91 

Brick,  10,  14,  17,  23 

Building  stone,  25 

Butt  joints,  118 

Cantilever  beams,  56,  63,  68,  107 
Cast  iron,  5,  8,  10,  11,  14,  17,  26 
Cement,  34 
Center  of  gravity,  47,  52 

moments,  38 
Cold-bend  test,  29 
Columns,  77-91,  131 
Combined  stresses,  146-153 
Compound  bars,  129 
Compression,  1,  9,  77 


Compressive  strength,  10 
tests,  19 

Concrete,  34,  127 

beams,  134,  137 
columns,  131 
reinforced,  126 

Deflection  of  beams,  106,  108 
Deformations,  2,  103 
Design  of  beams,  67,  140 

columns,  85,  131 
Direct  stresses,  1 
Ductility,  8 

Economic  construction,  14 
Elastic  deformations,  103 
limit,  4,  5,  19 
resilience,  154,  156 
strength,  1,  5 
Elongation,  2,  7,  103 

Factor  of  safety,  6,  14 

Granite,  25 
Gyration,  radius  of,  79 

Hemlock,  22 
Hoops,  121 
Hollow  shafts,  101 
Horse-power,  98 
Hydraulic  cement,  34 


167 


168 


INDEX 


I  beams,  47,  72 
Impact,  161 
Investigation,  13,  66 

Lap  joints,  115 
Limestone,  25 

Modulus  of  elasticity,  104 
resilience,  156 
rupture,  124 

Moments  for  beams,  37,  41,  62 
of  inertia,  45,  50,  53, 
57 

Mortar,  34 

Neutral  axis,  44,  87,  137,  139 

surface,  44,  57 
Nickel  steel,  33 

Pig  iron,  26 

Pine,  22 

Pipes,  114,  117 

Polar  moments  of  inertia,  94 

Radius  of  gyration,  79 

Rankine's  formula,  81 

Railroad  rails,  32 

Reactions  of  beams,  39,  59 

Reinforced  concrete,  126-145 

Resilience,  154 

of  bars,  155 
of  beams,  157 

Resisting  moment,  43,  61 

Restrained  beams,  110 

Riveted  joints,  115,  118 

Round  shafts,  96,  99 

Ropes,  35 

Sandstone,  25 


Safe  loads  for  beams,  64,  68 

columns,  83 
Shaft  couplings,  122 
Shafts,  92-102,  112,  123,  150 
Shear,  1,  2,  12,  22,  58 
Shortening,  2,  44 
Shrinkage  of  hoops,  121 
Simple  beams,  39,  42,  56,  58 
Square  shafts,  100 
Steam  pipes,  113,  117 
Steel,  5,  8,  10,  11,  14,  17,  31 
Steel  castings,  33 
Steel  I  beams,  72 
Strength,  elastic,  4,  69 

ultimate,  5-15 
Stone,  10,  14,  17,  25 
Stress,  1,  7,  13 
Structural  steel,  33 
Struts,  77-91 
Sudden  loads,  13,  159 

T  sections,  52 

Tables: 

Compressive  strengths,  10 
Constant  for  columns,  82 
Elastic  limits,  5 
Factors  of  safety,  14 
Modulus  of  elasticity,  104 
Modulus  of  rupture,  124 
Shearing  strengths,  11 
Steel  I  beams,  72 
Strength  of  timber,  22 
Tensile  strengths,  8 
Weights  of  materials,  17 

Temperature  stresses,  120 

Tensile  tests,  12,  18 

Tension,  1,  7,  8,  151 

Testing  machines,  18,  20 

Test  specimens,  12,  20 

Timber,  5,  8,  10,  11,  14,  17,  21 


INDEX 


169 


Tires,  121 

Torsion,  92-102,  149,  150 

Trap  rock,  25 

Twist  in  shafts,  112 

Ultimate  elongation,  8,  30 

resilience,  155 

strength,  5,  8,  10,  11, 

15,  163 
Unit-stress,  1,  3,  12 


Vertical  shear,  59 

Water  pipes,  114 
Weight  of  bars,  18 

concrete,  126 

materials,  17 
Wire,  30 
Work,  91,  105 

Working  unit-stress,  12,  164 
Wrought  iron,  5,  8, 10,  11,  14,  17, 
28 


This  book  is  DUE  on  the  last  date  stamped  below 


OCT  1  4  193* 

2  8  1932 
APR  il  19321 


1  1  1934 
NOV  1     193? 


MAY  1 
«B1- 


iwV   I  4 
5  193* 


Form  L-9-10m-5,'28 


332 


405 
Ivv55s 


UC  SOUTHERN  REGIONAL  LIBRARY  FACILITY 


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